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The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials
106 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + β¦ H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + β¦ (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons β’C β C - C β C - C β C - C - Cβ’ + β¦ H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene.
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It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials
106 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + β¦ H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + β¦ (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons β’C β C - C β C - C β C - C - Cβ’ + β¦ H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C β C - C β C - C β C - C β C- + β¦
107 H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C β C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC).
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Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials
106 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + β¦ H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + β¦ (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons β’C β C - C β C - C β C - C - Cβ’ + β¦ H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C β C - C β C - C β C - C β C- + β¦
107 H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C β C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic.
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During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + β¦ H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + β¦ (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons β’C β C - C β C - C β C - C - Cβ’ + β¦ H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C β C - C β C - C β C - C β C- + β¦
107 H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C β C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
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Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C β C - C β C - C β C - C β C- + β¦
107 H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C β C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + β¦ H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + β¦
108 (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons β’ C β C - C β C - C β C - C - C β’ + β¦ H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene.
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It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + β¦ H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + β¦
108 (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons β’ C β C - C β C - C β C - C - C β’ + β¦ H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C β C - C β C - C β C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C β C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)
109 104 The commercial name of polyphenylethene is polystyrene.
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PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + β¦ H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + β¦
108 (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons β’ C β C - C β C - C β C - C - C β’ + β¦ H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C β C - C β C - C β C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C β C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)
109 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic.
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During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + β¦ H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + β¦
108 (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons β’ C β C - C β C - C β C - C - C β’ + β¦ H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C β C - C β C - C β C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C β C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)
109 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators.
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Polyphenylethene molecule can be represented as: H H H H H H H H - C β C - C β C - C β C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C β C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)
109 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 4.Formation of Polypropene Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
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It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 4.Formation of Polypropene Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + β¦ H CH3 H CH3 H CH3 H CH3 propene + propene + propene + propene + β¦ (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ H CH3 H CH3 H CH3 H CH3 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons β’ C β C - C β C - C β C - C - C β’ + β¦ H CH3 H CH3 H CH3 H CH3 Lone pair of electrons can be used to join more monomers to form longer propene. propene molecule can be represented as:
110 H H H H H H H H - C β C - C β C - C β C - C - C - H CH3 H CH3 H CH3 H CH3 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C β C )n H CH3 Examples Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760 Substituting 4760 = 108.1818 =>108 propene molecules(whole number) 44 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic.
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During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + β¦ H CH3 H CH3 H CH3 H CH3 propene + propene + propene + propene + β¦ (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ H CH3 H CH3 H CH3 H CH3 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons β’ C β C - C β C - C β C - C - C β’ + β¦ H CH3 H CH3 H CH3 H CH3 Lone pair of electrons can be used to join more monomers to form longer propene. propene molecule can be represented as:
110 H H H H H H H H - C β C - C β C - C β C - C - C - H CH3 H CH3 H CH3 H CH3 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C β C )n H CH3 Examples Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760 Substituting 4760 = 108.1818 =>108 propene molecules(whole number) 44 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
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It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F
111 C = C + C = C + C = C + C = C + β¦ F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + β¦ (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons β’C β C - C β C - C β C - C - Cβ’ + β¦ F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene.
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Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F
111 C = C + C = C + C = C + C = C + β¦ F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + β¦ (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons β’C β C - C β C - C β C - C - Cβ’ + β¦ F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C β C - C β C - C β C - C β C- + β¦ F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C β C )n F F Examples
112 Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E).
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(ii)ceiling tiles (iii)clothe linings 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F
111 C = C + C = C + C = C + C = C + β¦ F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + β¦ (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons β’C β C - C β C - C β C - C - Cβ’ + β¦ F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C β C - C β C - C β C - C β C- + β¦ F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C β C )n F F Examples
112 Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic.
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During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F
111 C = C + C = C + C = C + C = C + β¦ F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + β¦ (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F β’C β Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β’C - Cβ’ + β¦ F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons β’C β C - C β C - C β C - C - Cβ’ + β¦ F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C β C - C β C - C β C - C β C- + β¦ F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C β C )n F F Examples
112 Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 5.Formation of rubber from Latex Natural rubber is obtained from rubber trees.
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polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C β C - C β C - C β C - C β C- + β¦ F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C β C )n F F Examples
112 Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 5.Formation of rubber from Latex Natural rubber is obtained from rubber trees. During harvesting an incision is made on the rubber tree to produce a milky white substance called latex. Latex is a mixture of rubber and lots of water. The latex is then added an acid to coagulate the rubber. Natural rubber is a polymer of 2-methylbut-1,3-diene ; H CH3 H H CH2=C (CH3) CH = CH2 H - C = C β C = C - H During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon β2β thus; H CH3 H H H CH3 H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H CH3 H H -(- C - C = C - C -)n-
113 H H Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher.
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The latex is then added an acid to coagulate the rubber. Natural rubber is a polymer of 2-methylbut-1,3-diene ; H CH3 H H CH2=C (CH3) CH = CH2 H - C = C β C = C - H During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon β2β thus; H CH3 H H H CH3 H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H CH3 H H -(- C - C = C - C -)n-
113 H H Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher. During vulcanization the sulphur atoms form a cross link between chains of rubber molecules/polymers. This decreases the number of C=C double bonds in the polymer. H CH3 H H H CH3 H H - C - C - C - C - C - C - C - C - H S H H S H H CH3 S H H CH3 S H - C - C - C - C - C - C - C - C - H H H H H H Vulcanized rubber is used to make tyres, shoes and valves. 6.Formation of synthetic rubber Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot. Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ; H Cl H H CH2=C (Cl CH = CH2 H - C = C β C = C - H During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon β2β thus; H Cl H H H Cl H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; Sulphur atoms make cross link between polymers
114 H Cl H H -(- C - C = C - C -)n- H H Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber.
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6.Formation of synthetic rubber Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot. Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ; H Cl H H CH2=C (Cl CH = CH2 H - C = C β C = C - H During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon β2β thus; H Cl H H H Cl H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; Sulphur atoms make cross link between polymers
114 H Cl H H -(- C - C = C - C -)n- H H Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber. (b)Condensation polymerization Condensation polymerization is the process where two or more small monomers join together to form a larger molecule by elimination/removal of a simple molecule. (usually water). Condensation polymers acquire a different name from the monomers because the two monomers are two different compounds During condensation polymerization: (i)the two monomers are brought together by high pressure to reduce distance between them. (ii)monomers realign themselves at the functional group. (iii)from each functional group an element is removed so as to form simple molecule (of usually H2O/HCl) (iv)the two monomers join without the simple molecule of H2O/HCl Examples of condensation polymerization 1.Formation of Nylon-6,6 Method 1: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioic acid with hexan-1,6-diamine.Amines are a group of homologous series with a general formula R-NH2 and thus -NH2 as the functional group. During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
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(ii)monomers realign themselves at the functional group. (iii)from each functional group an element is removed so as to form simple molecule (of usually H2O/HCl) (iv)the two monomers join without the simple molecule of H2O/HCl Examples of condensation polymerization 1.Formation of Nylon-6,6 Method 1: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioic acid with hexan-1,6-diamine.Amines are a group of homologous series with a general formula R-NH2 and thus -NH2 as the functional group. During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O H H H- O - C β (CH2 ) 4 β C β O - H + H βN β (CH2) 6 β N β H
115 (iii)from each functional group an element is removed so as to form a molecule of H2O and the two monomers join at the linkage . O O H H H- O - C β (CH2 ) 4 β C β N β (CH2) 6 β N β H + H 2O . Polymer bond linkage Nylon-6,6 derive its name from the two monomers each with six carbon chain Method 2: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioyl dichloride with hexan-1,6-diamine. Hexan-1,6-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group. The R-OCl is formed when the βOHβ in R-OOH/alkanoic acid is replaced by Cl/chlorine/Halogen During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
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Polymer bond linkage Nylon-6,6 derive its name from the two monomers each with six carbon chain Method 2: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioyl dichloride with hexan-1,6-diamine. Hexan-1,6-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group. The R-OCl is formed when the βOHβ in R-OOH/alkanoic acid is replaced by Cl/chlorine/Halogen During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O H H Cl - C β (CH2 ) 4 β C β Cl + H βN β (CH2) 6 β N β H (iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage . O O H H Cl - C β (CH2 ) 4 β C β N β (CH2) 6 β N β H + HCl . Polymer bond linkage The two monomers each has six carbon chain hence the name βnylon-6,6β The commercial name of Nylon-6,6 is Nylon It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and carpets. 2.Formation of Terylene
116 Method 1: Terylene can be made from the condensation polymerization of ethan1,2-diol with benzene-1,4-dicarboxylic acid. Benzene-1,4-dicarboxylic acid a group of homologous series with a general formula R-COOH where R is a ring of six carbon atom called Benzene ring .The functional group is -COOH. During the formation of Terylene: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
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2.Formation of Terylene
116 Method 1: Terylene can be made from the condensation polymerization of ethan1,2-diol with benzene-1,4-dicarboxylic acid. Benzene-1,4-dicarboxylic acid a group of homologous series with a general formula R-COOH where R is a ring of six carbon atom called Benzene ring .The functional group is -COOH. During the formation of Terylene: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O H- O - C β C6H5 β C β O - H + H βO β CH2 CH2 β O β H (iii)from each functional group an element is removed so as to form a molecule of H2O and the two monomers join at the linkage . O O H- O - C β C6H5 β C β O β (CH2) 6 β N β H + H 2O . Polymer bond linkage of terylene
117 Method 2: Terylene can be made from the condensation polymerization of benzene-1,4-dioyl dichloride with ethan-1,2-diol. Benzene-1,4-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group and R as a benzene ring. The R-OCl is formed when the βOHβ in R-OOH is replaced by Cl/chlorine/Halogen During the formation of Terylene (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O Cl - C β C5H5 β C β Cl + H βO β CH2 CH2 β O - H (iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage . O O Cl - C β C5H5 β C β O β CH2 CH2 β O β H + HCl . Polymer bond linkage of terylene
118 The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic.
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O O Cl - C β C5H5 β C β Cl + H βO β CH2 CH2 β O - H (iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage . O O Cl - C β C5H5 β C β O β CH2 CH2 β O β H + HCl . Polymer bond linkage of terylene
118 The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and sails and plastic model kits. 119 Practice questions Organic chemistry 1. A student mixed equal volumes of Ethanol and butanoic acid. He added a few drops of concentrated Sulphuric (VI) acid and warmed the mixture (i) Name and write the formula of the main products Nameβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. Formulaβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (ii) Which homologous series does the product named in (i) above belong? 2. The structure of the monomer phenyl ethene is given below:- a) Give the structure of the polymer formed when four of the monomers are added together b) Give the name of the polymer formed in (a) above 3. Explain the environmental effects of burning plastics in air as a disposal method 4. Write chemical equation to represent the effect of heat on ammonium carbonate 5. Sodium octadecanoate has a chemical formula CH3(CH2)6 COO-Na+, which is used as soap. HC = CH2 O
120 Explain why a lot of soap is needed when washing with hard water 6. A natural polymer is made up of the monomer: (a) Write the structural formula of the repeat unit of the polymer (b) When 5.0 x 10-5 moles of the polymer were hydrolysed, 0.515g of the monomer were obtained. Determine the number of the monomer molecules in this polymer. (C = 12; H = 1; N = 14; O =16) 7. The formula below represents active ingredients of two cleansing agents A and B O CH3CH2CH C OH
121 Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain
122 8.
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(C = 12; H = 1; N = 14; O =16) 7. The formula below represents active ingredients of two cleansing agents A and B O CH3CH2CH C OH
121 Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain
122 8. Study the polymer below and use it to answer the questions that follow: (a) Give the name of the monomer and draw its structures (b) Identify the type of polymerization that takes place (c) State one advantage of synthetic polymers 9. Ethanol and Pentane are miscible liquids. Explain how water can be used to separate a mixture of ethanol and pentane 10. (a) What is absolute ethanol? GLUCOSE SOLUTION CRUDE ETHANOL 95% ETHANOL ABSOLUTE ETHANOL G H H H H H C C C C
123 (b) State two conditions required for process G to take place efficiently 11. (a) (i) The table below shows the volume of oxygen obtained per unit time when hydrogen peroxide was decomposed in the presence of manganese (IV) Oxide. Use it to answer the questions that follow:- Time in seconds Volume of Oxygen evolved (cm3) 0 30 60 90 120 150 180 210 240 270 300 0 10 19 27 34 38 43 45 45 45 45 (i) Plot a graph of volume of oxygen gas against time
124 (ii) Determine the rate of reaction at time 156 seconds (iii) From the graph, find the time taken for 18cm3 of oxygen to be produced (iv) Write a chemical equation to show how hydrogen peroxide decomposes in the presence of manganese (IV) Oxide (b) The diagram below shows how a Leβclanche (Dry cell) appears:- (i) What is the function of MnO2 in the cell above? (ii) Write the equation of a reaction that occurs at the cathode
125 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65) 12.
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(a) (i) The table below shows the volume of oxygen obtained per unit time when hydrogen peroxide was decomposed in the presence of manganese (IV) Oxide. Use it to answer the questions that follow:- Time in seconds Volume of Oxygen evolved (cm3) 0 30 60 90 120 150 180 210 240 270 300 0 10 19 27 34 38 43 45 45 45 45 (i) Plot a graph of volume of oxygen gas against time
124 (ii) Determine the rate of reaction at time 156 seconds (iii) From the graph, find the time taken for 18cm3 of oxygen to be produced (iv) Write a chemical equation to show how hydrogen peroxide decomposes in the presence of manganese (IV) Oxide (b) The diagram below shows how a Leβclanche (Dry cell) appears:- (i) What is the function of MnO2 in the cell above? (ii) Write the equation of a reaction that occurs at the cathode
125 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65) 12. (a) Give the IUPAC names of the following compounds: (i) CH3COOCH2CH3 * (ii) (b) The structure below shows some reactions starting with ethanol. Study it and answer the questions that follow: CH2 = C β CHCH3 CH3COOH CH3COONa CH3CH2OH CH2=CH2 CH3CH3 CH2 CH2 n S P CH4 T Na Metal Compound U Step II Step I Step III CH3COOH Reagent R NaOH(aq) Heat Excess Cl2/U V
126 (i) Write the formula of the organic compounds P and S * (ii) Name the type of reaction, the reagent(s) and condition for the reactions in the following steps :- (I) Step I * (II) Step II * (III) Step III * (iii) Name reagent R β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ * (iv) Draw the structural formula of T and give its name * (v) (I) Name compound Uβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..
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(ii) Write the equation of a reaction that occurs at the cathode
125 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65) 12. (a) Give the IUPAC names of the following compounds: (i) CH3COOCH2CH3 * (ii) (b) The structure below shows some reactions starting with ethanol. Study it and answer the questions that follow: CH2 = C β CHCH3 CH3COOH CH3COONa CH3CH2OH CH2=CH2 CH3CH3 CH2 CH2 n S P CH4 T Na Metal Compound U Step II Step I Step III CH3COOH Reagent R NaOH(aq) Heat Excess Cl2/U V
126 (i) Write the formula of the organic compounds P and S * (ii) Name the type of reaction, the reagent(s) and condition for the reactions in the following steps :- (I) Step I * (II) Step II * (III) Step III * (iii) Name reagent R β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ * (iv) Draw the structural formula of T and give its name * (v) (I) Name compound Uβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (II) If the relative molecular mass of U is 42000, determine the value of n (C=12, H=1) (c) State why C2H4 burns with a more smoky flame than C2H6 * 13. a) State two factors that affect the properties of a polymer b) Name the compound with the formula below : CH3CH2CH2ONa c) Study the scheme below and use it to answer the questions that follow:- CH3CH2CH3 P Step CH3CH = CH2 Step CH3CH2CH2OH Step R H H C β C n K
127 i) Name the following compounds:- I. Product T β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ II.
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(II) If the relative molecular mass of U is 42000, determine the value of n (C=12, H=1) (c) State why C2H4 burns with a more smoky flame than C2H6 * 13. a) State two factors that affect the properties of a polymer b) Name the compound with the formula below : CH3CH2CH2ONa c) Study the scheme below and use it to answer the questions that follow:- CH3CH2CH3 P Step CH3CH = CH2 Step CH3CH2CH2OH Step R H H C β C n K
127 i) Name the following compounds:- I. Product T β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ II. K β¦β¦β¦ ii) State one common physical property of substance G iii) State the type of reaction that occurred in step J
128 iv) Give one use of substance K v) Write an equation for the combustion of compound P vi) Explain how compounds CH3CH2COOH and CH3CH2CH2OH can be distinguished chemically vii) If a polymer K has relative molecular mass of 12,600, calculate the value of n (H=1 C =12) 14. Study the scheme given below and answer the questions that follow:- H2 (g) Ni High temp Polymer Q Polymerization Compound P CH3CH2CH3 CH3CH2CH2ONa + H2 Na(s) Propan-l-ol Step I Propylethanoate CH3CH2COOH Solution T + CO2 (g) Step III Na2CO3(aq) Conc. H2SO4 180oC Step II
129 (a) (i) Name compound P β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ (ii) Write an equation for the reaction between CH3CH2COOH and Na2CO3 (b) State one use of polymer Q (c) Name one oxidising agent that can be used in step II β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (d) A sample of polymer Q is found to have a molecular mass of 4200. Determine the number of monomers in the polymer (H = 1, C = 12) (e) Name the type of reaction in step I β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (f) State one industrial application of step III (g)State how burning can be used to distinguish between propane and propyne.
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(d) A sample of polymer Q is found to have a molecular mass of 4200. Determine the number of monomers in the polymer (H = 1, C = 12) (e) Name the type of reaction in step I β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (f) State one industrial application of step III (g)State how burning can be used to distinguish between propane and propyne. Explain your answer (h) 1000cm3 of ethene (C2H4) burnt in oxygen to produce Carbon (II) Oxide and water vapour. 130 Calculate the minimum volume of air needed for the complete combustion of ethene (Air contains 20% by volume of oxygen) 15. (a) Study the schematic diagram below and answer the questions that follow:- (i) Identify the following: Substance Q .............................................................................................................. Substance R............................................................................................................... Gas P.......................................................................................................................... CH3CH2COOCH2CH2CH3 CH3CHCH2 CH3CH2CH2ONa + Gas P CH3CH2CH2OH X V HCl Step 5 Step 1 R Na H+ Step 3 Q + H2O MnO4 Step 4Ni H2
131 (ii) Name: Step 1................................................................................................. Step 4................................................................................................. (iii) Draw the structural formula of the major product of step 5 (iv) State the condition and reagent in step 3 16. Study the flow chart below and answer the questions that follow (a) (i) Name the following organic compounds: Mβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..β¦β¦.. Lβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. M KMnO4/H+ CH2CH2 Ethyl Ethanoate CH2CH2OH L J K CO2 (g) STEP 2 Reagent Q St3KMnO4/H+(aq) Ni/H2(g) Step 4 Reagent P
132 Products C2H5COONa Step V ClbiCH οΊCH Step I Step II CH2 = CH2 Step III C2H6 Step IV + Heat (ii) Name the process in step: Step 2 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦. Step 4 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦ (iii) Identify the reagent P and Q (iv) Write an equation for the reaction between CH3CH2CH2OH and sodium 17.
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Lβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. M KMnO4/H+ CH2CH2 Ethyl Ethanoate CH2CH2OH L J K CO2 (g) STEP 2 Reagent Q St3KMnO4/H+(aq) Ni/H2(g) Step 4 Reagent P
132 Products C2H5COONa Step V ClbiCH οΊCH Step I Step II CH2 = CH2 Step III C2H6 Step IV + Heat (ii) Name the process in step: Step 2 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦. Step 4 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦ (iii) Identify the reagent P and Q (iv) Write an equation for the reaction between CH3CH2CH2OH and sodium 17. a) Give the names of the following compounds: i) CH3CH2CH2CH2OH β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ ii) CH3CH2COOH β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ iii) CH3C β O- CH2CH3 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ 18. Study the scheme given below and answer the questions that follow;
133 n i) Name the reagents used in: Step I: β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ Step II β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ Step III β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ ii) Write an equation to show products formed for the complete combustion of CH = CH iii) Explain one disadvantage of continued use of items made form the compound formed in step III 19. A hydrated salt has the following composition by mass. Iron 20.2 %, oxygen 23.0%, sulphur 11.5%, water 45.3% i) Determine the formula of the hydrated salt (Fe=56, S=32, O=16, H=11) ii) 6.95g of the hydrated salt in c(i) above were dissolved in distilled water and the total volume made to 250cm3 of solution. Calculate the concentration of the resulting salt solution in moles per litre. (Given that the molecula mass of the salt is 278) 20. Write an equation to show products formed for the complete combustion of CH = CH
134 iii) Explain one disadvantage of continued use of items made form the compound formed in step III
135 21.
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Calculate the concentration of the resulting salt solution in moles per litre. (Given that the molecula mass of the salt is 278) 20. Write an equation to show products formed for the complete combustion of CH = CH
134 iii) Explain one disadvantage of continued use of items made form the compound formed in step III
135 21. Give the IUPAC name for each of the following organic compounds; i) CH3 - CH - CH2 - CH3 OH ii)CH3 β CH β CH2 β CH2 - CH3 C2H5 iii)CH3COOCH2CH2CH3 22. The structure below represents a cleansing agent. O R β S β O-Na+ O a) State the type of cleansing agent represented above b) State one advantage and one disadvantage of using the above cleansing agent. 23. The structure below shows part of polymer .Use it to answer the questions that follow. CH3 CH3 CH3 ο― ο― ο―
136 β CH - CH2 β CH- CH2 - CH β CH2 β a) Derive the structure of the monomer b) Name the type of polymerization represented above 24. The flow chart below represents a series of reactions starting with ethanoic acid:- (a) Identify substances A and B (b) Name the process I 25. a) Write an equation showing how ammonium nitrate may be prepared starting with ammonia gas (b) Calculate the maximum mass of ammonium nitrate that can be prepared using 5.3kg of ammonia (H=1, N=14, O=16) 26. (a) What is meant by the term, esterification? Ethanol B Ethanoic acid Na2CO3 Salt A + CO2 + H2O
137 (b) Draw the structural formulae of two compounds that may be reacted to form ethylpropanoate 27. (a) Draw the structure of pentanoic acid (b) Draw the structure and give the name of the organic compound formed when ethanol reacts with pentanoic acid in presence of concentrated sulphuric acid
138 28. The scheme below shows some reactions starting with ethanol.
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Ethanol B Ethanoic acid Na2CO3 Salt A + CO2 + H2O
137 (b) Draw the structural formulae of two compounds that may be reacted to form ethylpropanoate 27. (a) Draw the structure of pentanoic acid (b) Draw the structure and give the name of the organic compound formed when ethanol reacts with pentanoic acid in presence of concentrated sulphuric acid
138 28. The scheme below shows some reactions starting with ethanol. Study it and answer the questions that follow:- (i) Name and draw the structure of substance Q Q P CH3COONa Ethanol CH3CH2ONa C H2SO4(l) Cr2O7(aq) / H+(aq) Na(s) Step 2 Step 4 CH3CH2OH/H2SO4 Step 3 2-
139 (ii) Give the names of the reactions that take place in steps 2 and 4 (iii) What reagent is necessary for reaction that takes place in step 3 29. Substances A and B are represented by the formulae ROH and RCOOH respectively. They belong to two different homologous series of organic compounds. If both A and B react with potassium metal: (a) Name the common product produced by both (b) State the observation made when each of the samples A and B are reacted with sodium hydrogen carbonate (i) A (ii) B 30. Below are structures of particles. Use it to answer questions that follow. In each case only electrons in the outermost energy level are shown key P = Proton N = Neutron X = Electron W U V 19P Z Y
140 (a) Identify the particle which is an anion 31. Plastics and rubber are extensively used to cover electrical wires. (a) What term is used to describe plastic and rubbers used in this way? (b) Explain why plastics and rubbers are used this way
141 32. The scheme below represents the manufacture of a cleaning agent X (a) Draw the structure of X and state the type of cleaning agent to which X belong (b) State one disadvantage of using X as a cleaning agent 33. Y grams of a radioactive isotope take 120days to decay to 3.5grams.
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(b) Explain why plastics and rubbers are used this way
141 32. The scheme below represents the manufacture of a cleaning agent X (a) Draw the structure of X and state the type of cleaning agent to which X belong (b) State one disadvantage of using X as a cleaning agent 33. Y grams of a radioactive isotope take 120days to decay to 3.5grams. The halflife period of the isotope is 20days (a) Find the initial mass of the isotope (b) Give one application of radioactivity in agriculture 34. The structure below represents a polymer. Study and answer the questions that follow:- R Conc. R Cleaning agent X H H -C β C - n
142 (i) Name the polymer above.................................................................................. (ii) Determine the value of n if giant molecule had relative molecular mass of 4956 35. RCOO-Na+ and RCH2OSO3-Na+ are two types of cleansing agents; i) Name the class of cleansing agents to which each belongs ii) Which one of these agents in (i) above would be more suitable when washing with water from the Indian ocean. Explain iii) Both sulphur (IV) oxide and chlorine are used bleaching agents. Explain the difference in their bleaching properties 36. The formula given below represents a portion of a polymer H H H H C C C C O H O H n
143 (a) Give the name of the polymer (b) Draw the structure of the monomer used to manufacture the polymer
15.0.0 NITROGEN AND IT'S COMPOUNDS (30 LESSONS) Introduction to oxides of nitrogen Nirogen has a position in second period of group V in the modern periodic table. It has molecular formula N2. It has atomic number 7 and atomic weight 14.08 and its electronic configuration of 2,5. Besides combining with hydrogen and forming NH3 , nitrogen combines with oxygen in different ratios and forms five different oxides.
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It has molecular formula N2. It has atomic number 7 and atomic weight 14.08 and its electronic configuration of 2,5. Besides combining with hydrogen and forming NH3 , nitrogen combines with oxygen in different ratios and forms five different oxides. The oxides of nitrogen and the details of the oxygen states of nitrogen and the N:O ratio can be presented in a tabular form as: Name Formula Oxidation state of N Ratio of N:O 1) Nitrogen Oxide or nitrous oxide NO +2 1:1 2) Nitrogen dioxide NO2 +4 1:2 3) Nitrous oxide (also called laughing gas) N2O +1 2:1 All of these oxides of nitrogen are gases, excepting NO and N2O the other oxides are brownish gases. Except the oxides of NO and N2O all the other oxides are acidic. NO and N2O are neutral. The other oxides are prepared in the laboratory using different methods characteristic to each oxide. Example of Oxides of Nitrogen (nitric Oxide - NO) Structure:
Laboratory Preparation: The oxide is prepared in the laboratory by treating the metallic copper with a moderately concentarted nitric acid (1:1) at room temperature. The reaction is given as : 2Cu + 8 HNO3 ------> Cu(NO3)2 + 2NO + 4H2O The gas is collected by downward displacement of water.The apparatus used is Wolfe's apparatus. The purification is done by absorbing the NO gas in frashly prepared ferrous sulphate solution. Ferrous sulphate absorbs all the NO gas and forms Fe(H2O)5 NO and the solution becomes brown. On heating this solution pure Nitric Oxide is obatined. Physical Properties: NO is not a combustible gas. At high temperature around 1000Β°C it decomposes into N2 and O2 . 2NO = N2 + O2 at high temperature From the equation above we can see that once the decomposition starts 50% O2 gets evolved and this O2 supports combustion thus making the reaction more violent.
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Physical Properties: NO is not a combustible gas. At high temperature around 1000Β°C it decomposes into N2 and O2 . 2NO = N2 + O2 at high temperature From the equation above we can see that once the decomposition starts 50% O2 gets evolved and this O2 supports combustion thus making the reaction more violent. Chemical properties: 1) NO acts as an oxidising agent, oxidising SO2 in presence of water to give H2SO4: SO2 + 2NO + H2O β H2SO4 + N2O 2) NO acts as a reducing agent, i) reducing an acidified solution of potassium permanganate (pink) to colorless manganous salt. 3KMnO4 + 6H2SO4 + 5NO2 β 3KHSO4 + 3MnSO4 + 2H2O + 5NO3
ii) It can also reduce aqueous solution of I2 to HI 3I2 + 2NO + 4H2O β 2HNO3 + 6HI 3) With halogens NO can form addition compounds as 2NO + Cl2 β 2NOCl (NOCl is nitrosyl chloride) It reacts in the same way with flourine and bromine. 4) With ferrous sulphate NO forms an addition compound as FeSO4 + 5H2O + NO = [Fe(H2O)5NO]SO4 penta aqua nitrosyl iron (II) sulphate This is the famous brown ring test used to identify the nitrate radical or the NO radical. Uses: NO is used to prepare nitric acid. Nitrogen Dioxide - NO2 Structure: Laboratory Preparation: In the laboratory NO2 is prepared by thermal decomposition of Pb(NO3)2. Thus 2Pb(NO3)2 β 2PbO + 4NO2 + O2 Care is taken to ensure the use of dried Pb(NO3)2 as hydrated nitrate salts on heating react violently and explode. Physical Properties: NO2 is a poisonous gas, main source being the exhaust of automobiles. 1) At room temperature it is a deep brown gas. 2) It does not support combustion. 3) It is not combustible. Chemical Properties: 1) With cold water NO2 reacts to give a mixture of HNO2 and HNO3 acid.
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2) It does not support combustion. 3) It is not combustible. Chemical Properties: 1) With cold water NO2 reacts to give a mixture of HNO2 and HNO3 acid. 2NO2 + H2O β HNO2 + HNO3 2) With hot water the reaction is 3NO2 + H2O β 2HNO3 + NO 3) Being acidic it reacts with bases as 2NO2 + KOH β KNO3 + H2O + KNO2 4) It is also a strong oxidising agent. H2S + NO2 β NO + H2S + S 5) With excess oxygen and water NO2 gives HNO3. 4NO2 + O2 + 2H2O β 4HNO3 6) It reacts with concentrated H2SO4 to give nitrosyl hydrogen sulphate 2NO2 + H2SO4 β SO2(OH)ONO + HNO3 Uses: NO2 is used as a fuel in rockets besides being used to prepare HNO3 . Nitrous Oxide - N2O (laughing Gas) Structure:
Laboratory Preparation: N2O can be prepared in the laboratory by heating NH4NO3 below 200Β°C to avoid explosion. Sometimes as a safety measure instead of directly using NH4NO3, a mixture of (NH4)2SO4 and NaNO3 are heated to give NH4NO3 which decomposes further to give N2O. NH4NO3 β N2O + H2O (endothermic reaction) Physical Properties: 1) N2O has a faint sweet smell and produces a tickling sensation on the neck when inhaled and makes people laugh hysterically. Excess of inhalation leads to unconsiousness. 2) Unlike other oxides of nitrogen, N2O supports combustion though it does not burn itself. Chemical Properties: 1) At very high temperature N2O decomposes to N2 and O2 2N2O β 2N2 + O2 If a glowing piece of Mg, Cu, or P is introduced in such an environment, these pieces burn brightly due to the O2 produced from decomposition of N2O. 2) With Sodium and potassium N2O reacts to give the corresponding peroxides liberating N2 in the process.
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2) Unlike other oxides of nitrogen, N2O supports combustion though it does not burn itself. Chemical Properties: 1) At very high temperature N2O decomposes to N2 and O2 2N2O β 2N2 + O2 If a glowing piece of Mg, Cu, or P is introduced in such an environment, these pieces burn brightly due to the O2 produced from decomposition of N2O. 2) With Sodium and potassium N2O reacts to give the corresponding peroxides liberating N2 in the process. 2N2O + 2Na β Na2O + 2N2. Na2O is sodium peroxide Uses: 1) It is used as propellent gas. 2)Used in combination with oxygen in the ratio N2O : O2 = 1:10 as a mild anaesthetic. 1 16.0.0 SULPHUR AND ITS COMPOUNDS (25 LESSONS) A.SULPHUR (S) Sulphur is an element in Group VI Group 16)of the Periodic table . It has atomic number 16 and electronic configuration 16 and valency 2 /divalent and thus forms the ion S2- A. Occurrence. Sulphur mainly occurs : (i) as free element in Texas and Louisiana in USA and Sicily in Italy. (ii)Hydrogen sulphide gas in active volcanic areas e.g. Olkaria near Naivasha in Kenya (iii)as copper pyrites(CuFeS2) ,Galena (PbS,Zinc blende(ZnS))and iron pyrites(FeS2) in other parts of the world. B. Extraction of Sulphur from Fraschs process Suphur occurs about 200 metres underground. The soil structure in these areas is usually weak and can easily cave in. Digging of tunnels is thus discouraged in trying to extract the mineral. Sulphur is extracted by drilling three concentric /round pipes of diameter of ratios 2:8: 18 centimeters. Superheated water at 170oC and 10atmosphere pressure is forced through the outermost pipe. The high pressures ensure the water remains as liquid at high temperatures instead of vapour of vapour /gas. The superheated water melts the sulphur because the melting point of sulphur is lower at about at about 115oC.
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Superheated water at 170oC and 10atmosphere pressure is forced through the outermost pipe. The high pressures ensure the water remains as liquid at high temperatures instead of vapour of vapour /gas. The superheated water melts the sulphur because the melting point of sulphur is lower at about at about 115oC. A compressed air at 15 atmospheres is forced /pumped through the innermost pipe. The hot air forces the molten sulphur up the middle pipe where it is collected and solidifies in a large tank. It is about 99% pure. 2 Diagram showing extraction of Sulphur from Fraschs Process C. Allotropes of Sulphur. 1. Sulphur exist as two crystalline allotropic forms: (i)Rhombic sulphur (ii)Monoclinic sulphur Rhombic sulphur Monoclinic sulphur Bright yellow crystalline solid Has a melting point of 113oC Has a density of 2.06gcm-3 Stable below 96oC Has octahedral structure Pale yellow crystalline solid Has a melting point of 119oC Has a density of 1.96gcm-3 Stable above 96oC Has a needle-like structure Rhombic sulphur and Monoclinic sulphur have a transition temperature of 96oC.This is the temperature at which one allotrope changes to the other. βSketch of Octahedral structure ofRhombic sulphurSulphuratomsvalent bondNan-der-waalsforcesu Covalentbonds Sketch of the needle-like structure of monoclinic sulphur
4 2. Sulphur exists in non-crystalline forms as: (i)Plastic sulphur- Plastic sulphur is prepared from heating powdered sulphur to boil then pouring a thin continuous stream in a beaker with cold water. A long thin elastic yellow thread of plastic sulphur is formed .If left for long it turn to bright yellow crystalline rhombic sulphur. (ii)Colloidal sulphur- Colloidal sulphur is formed when sodium thiosulphate (Na2S2O3) is added hydrochloric acid to form a yellow precipitate. D. Heating Sulphur. A molecule of sulphur exists as puckered ring of eight atoms joined by covalent bonds as S8.
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D. Heating Sulphur. A molecule of sulphur exists as puckered ring of eight atoms joined by covalent bonds as S8. On heating the yellow sulphur powder melts at 113oC to clear amber liquid with low viscosity and thus flows easily. On further heating to 160oC the molten liquid darkens to a brown very viscous liquid that does not flow easily. This is because the S8 rings break into S8 chain that join together to form very long chains made of over 100000 atoms of Sulphur. The long chains entangle each other reducing their mobility /flow and hence increases their viscosity. On continued further heating to above 160oC, the viscous liquid darkens but becomes more mobile/flows easily and thus less viscous. This is because the long chains break to smaller/shorter chains. At 444oC, the liquid boils and forms brown vapour of a mixture of S8 ,S6 ,S2 molecules that solidifies to S8 ring of βflowers of sulphurβ on the cooler parts. Summary of changes on heating sulphur
5 Observation on heating Explanation/structure of Sulphur Solid sulphur Heat to 113oC Amber yellow liquid Heat to 160oC Liquid darkens Heat to 444oC Liquid boils to brown vapour Cool to room temperature Yellow sublimate (Flowers of Sulphur) Puckered S8 ring Puckered S8 ring in liquid form (low viscosity/flow easily) Puckered S8 ring break/opens then join to form long chains that entangle (very high viscosity/very low rate of flow) Mixture of S8 ,S6 ,S2 vapour Puckered S8 ring E. Physical and Chemical properties of Sulphur.(Questions) 1. State three physical properties unique to Sulphur Sulphur is a yellow solid, insoluble in water, soluble in carbon disulphide/tetrachloromethane/benzene, poor conductor of heat and electricity. It has a melting point of 115oC and a boiling point of 444oC. 2. Moist/damp/wet blue and red litmus papers were put in a gas jar containing air/oxygen. Burning sulphur was then lowered into the gas jar. State and explain the observation made.
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Moist/damp/wet blue and red litmus papers were put in a gas jar containing air/oxygen. Burning sulphur was then lowered into the gas jar. State and explain the observation made. Observations -Sulphur melts then burns with a blue flame Colourless gas produced that has a pungent smell Red litmus paper remains red. Blue litmus paper turns red. Explanation Sulphur burns in air and faster in Oxygen to form Sulphur(IV)Oxide gas and traces/small amount of Sulphur(VI)Oxide gas. Both oxides react with water to form the corresponding acidic solution i.e (i) Sulphur(IV)Oxide gas reacts with water to form sulphuric(IV)acid (ii) Sulphur(VI)Oxide gas reacts with water to form sulphuric(VI)acid Chemical equation
6 S(s) + O2(g) -> SO2(g) (Sulphur(IV)Oxide gas) 2S(s) + 3O2(g) -> 2SO3(g) (Sulphur(VI)Oxide gas traces) SO2(g) + H2O(l) -> H2 SO3 (aq) ( sulphuric(IV)acid) SO3(g) + H2O(l) -> H2 SO4 (aq) ( sulphuric(VI)acid). 3. Iron filings were put in a test tube containing powdered sulphur then heated on a Bunsen flame. Stop heating when reaction starts. State and explain the observations made. Test the effects of a magnet on the mixture before and after heating. Explain. Observations Before heating, the magnet attracts iron filings leaving sulphur After heating, the magnet does not attract the mixture. After heating, a red glow is observed that continues even when heating is stopped.. Black solid is formed. Explanation Iron is attracted to a magnet because it is ferromagnetic. When a mixture of iron and sulphur is heated, the reaction is exothermic giving out heat energy that makes the mixture to continue glowing even after stopping heating. Black Iron(II)sulphide is formed which is a compound and thus not ferromagnetic. Chemical equation Fe(s) + S(s) -> FeS(s) (Exothermic reaction/ -βH) Heated powdered heavy metals combine with sulphur to form black sulphides.
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When a mixture of iron and sulphur is heated, the reaction is exothermic giving out heat energy that makes the mixture to continue glowing even after stopping heating. Black Iron(II)sulphide is formed which is a compound and thus not ferromagnetic. Chemical equation Fe(s) + S(s) -> FeS(s) (Exothermic reaction/ -βH) Heated powdered heavy metals combine with sulphur to form black sulphides. Cu(s) + S(s) -> CuS(s) Zn(s) + S(s) -> ZnS(s) Pb(s) + S(s) -> PbS(s) 4.The set up below show the reaction of sulphur on heated concentrated sulphuric(VI)acid. 7 (i)State and explain the observation made. Observation Yellow colour of sulphur fades Orange colour of potassium dichromate(VI)paper turns to green. Explanation Hot concentrated sulphuric(VI)acid oxidizes sulphur to sulphur (IV)oxide gas. The oxide is also reduced to water. Traces of sulphur (VI)oxide is formed. Chemical equation S(s) + 3H2 SO4 (l) -> 3SO2(g) + 3H2O(l) +SO3(g) Sulphur (IV)oxide gas turns Orange potassium dichromate(VI)paper to green. (ii)State and explain the observation made if concentrated sulphuric (VI) acid is replaced with concentrated Nitric (V) acid in the above set up. Observation Yellow colour of sulphur fades Colurless solution formed Brown fumes/gas produced. Explanation Hot concentrated Nitric(V)acid oxidizes sulphur to sulphuric (VI)acid. The Nitric (V) acid is reduced to brown nitrogen(IV)oxide gas. Chemical equation S(s) + 6HNO3 (l) -> 6NO2(g) + 2H2O(l) +H2SO4 (l) NB: Hydrochloric acid is a weaker oxidizing agent and thus cannot oxidize sulphur like the other mineral acids. 5.State three main uses of sulphur . Sulphur is mainly used in: (i)Contact process for the manufacture/industrial/large scale production of concentrated sulphuric(VI)acid. (ii)Vulcanization of rubber to make it harder, tougher, stronger, and more durable.
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5.State three main uses of sulphur . Sulphur is mainly used in: (i)Contact process for the manufacture/industrial/large scale production of concentrated sulphuric(VI)acid. (ii)Vulcanization of rubber to make it harder, tougher, stronger, and more durable. (iii)Making gun powder and match stick heads (iv) As ointments to treat fungal infections
8 6. Revision Practice The diagram below represents the extraction of sulphur by Fraschs process. Use it to answer the questions that follow. (a)Name the substances that passes through: M Superheated water at 170oC and 10 atmosphere pressure L Hot compressed air N Molten sulphur (b)What is the purpose of the substance that passes through L and M? M- Superheated water at 170oC and 10 atmosphere pressure is used to melt the sulphur L- Hot compressed air is used to force up the molten sulphur. (c) The properties of the two main allotropes of sulphur represented by letters A and B are given in the table below. Use it to answer the questions that follow. A B Appearance Bright yellow Pale yellow Density(gcm-3) 1.93 2.08 Melting point(oC) 119 113 Stability Above 96oC Below 96oC I.What are allotropes? Different forms of the same element existing at the same temperature and pressure without change of state. II. Identify allotrope: A. Monoclinic sulphur B . Rhombic sulphur III. State two main uses of sulphur.
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Monoclinic sulphur B . Rhombic sulphur III. State two main uses of sulphur. -Manufacture of sulphuric(VI)acid -as fungicide -in vulcanization of rubber to make it harder/tougher/ stronger -manufacture of dyes /fibres L M N
9 (d)Calculate the volume of sulphur (IV)oxide produced when 0.4 g of sulphur is completely burnt in excess air.(S = 32.0 ,I mole of a gas occupies 24 dm3 at room temperature) Chemical equation S(s) + O2(g) -> SO2(g) Mole ratio S: SO2 = 1:1 Method 1 32.0 g of sulphur -> 24 dm3 of SO2(g) 0.4 g of sulphur -> 0.4 g x 24 dm3 = 0.3 dm3 32.0 g Method 2 Moles of sulphur used = Mass of sulphur => 0.4 = 0.0125 moles Molar mass of sulphur 32 Moles of sulphur used = Moles of sulphur(IV)oxide used=>0.0125 moles Volume of sulphur(IV)oxide used = Moles of sulphur(IV)oxide x volume of one mole of gas =>0.0125 moles x 24 dm3 = 0.3 dm3
10 B.COMPOUNDS OF SULPHUR The following are the main compounds of sulphur: (i) Sulphur(IV)oxide (ii) Sulphur(VI)oxide . (iii) Sulphuric(VI)acid (iv) Hydrogen Sulphide (v) Sulphate(IV)/SO32- and Sulphate(VI)/ SO42- salts (i) Sulphur(IV)oxide(SO2) (a) Occurrence Sulphur (IV)oxide is found in volcanic areas as a gas or dissolved in water from geysersand hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya. (b) School laboratory preparation In a Chemistry school laboratory Sulphur (IV)oxide is prepared from the reaction of Method 1:Using Copper and Sulphuric(VI)acid.
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(iii) Sulphuric(VI)acid (iv) Hydrogen Sulphide (v) Sulphate(IV)/SO32- and Sulphate(VI)/ SO42- salts (i) Sulphur(IV)oxide(SO2) (a) Occurrence Sulphur (IV)oxide is found in volcanic areas as a gas or dissolved in water from geysersand hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya. (b) School laboratory preparation In a Chemistry school laboratory Sulphur (IV)oxide is prepared from the reaction of Method 1:Using Copper and Sulphuric(VI)acid. Method 2:Using Sodium Sulphate(IV) and hydrochloric acid. 11 (c)Properties of Sulphur(IV)oxide(Questions) 1. Write the equations for the reaction for the formation of sulphur (IV)oxide using: (i)Method 1 Cu(s) + 2H2SO4(l) -> CuSO4(aq) + SO2(g) + 2H2O(l) Zn(s) + 2H2SO4(l) -> ZnSO4(aq) + SO2(g) + 2H2O(l) Mg(s) + 2H2SO4(l) -> MgSO4(aq) + SO2(g) + 2H2O(l) Fe(s) + 2H2SO4(l) -> FeSO4(aq) + SO2(g) + 2H2O(l) Calcium ,Lead and Barium will form insoluble sulphate(VI)salts that will cover unreacted metals stopping further reaction thus producing very small amount/quantity of sulphur (IV)oxide gas.
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Method 2:Using Sodium Sulphate(IV) and hydrochloric acid. 11 (c)Properties of Sulphur(IV)oxide(Questions) 1. Write the equations for the reaction for the formation of sulphur (IV)oxide using: (i)Method 1 Cu(s) + 2H2SO4(l) -> CuSO4(aq) + SO2(g) + 2H2O(l) Zn(s) + 2H2SO4(l) -> ZnSO4(aq) + SO2(g) + 2H2O(l) Mg(s) + 2H2SO4(l) -> MgSO4(aq) + SO2(g) + 2H2O(l) Fe(s) + 2H2SO4(l) -> FeSO4(aq) + SO2(g) + 2H2O(l) Calcium ,Lead and Barium will form insoluble sulphate(VI)salts that will cover unreacted metals stopping further reaction thus producing very small amount/quantity of sulphur (IV)oxide gas. (ii)Method 2 Na2SO3(aq) + HCl(aq) -> NaCl(aq ) + SO2(g) + 2H2O(l) K2SO3(aq) + HCl(aq) -> KCl(aq ) + SO2(g) + 2H2O(l) BaSO3(s) + 2HCl(aq) -> BaCl2(aq ) + SO2(g) + H2O(l) CaSO3(s) + 2HCl(aq) -> CaCl2(aq ) + SO2(g) + H2O(l) PbSO3(s) + 2HCl(aq) -> PbCl2(s ) + SO2(g) + H2O(l)
12 Lead(II)chloride is soluble on heating thus reactants should be heated to prevent it coating/covering unreacted PbSO3(s) 2.State the physical properties unique to sulphur (IV)oxide gas. Sulphur (IV)oxide gas is a colourless gas with a pungent irritating and choking smell which liquidifies easily. It is about two times denser than air. 3. The diagram below show the solubility of sulphur (IV)oxide gas. Explain.
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3. The diagram below show the solubility of sulphur (IV)oxide gas. Explain. Sulphur(IV) oxide is very soluble in water. One drop of water dissolves all the Sulphur (IV) oxide in the flask leaving a vacuum. If the clip is removed, atmospheric pressure forces the water up through the narrow tube to form a fountain to occupy the vacuum. An acidic solution of sulphuric (IV)acid is formed which turns litmus solution red. Chemical equation SO2(g) + H2O(l) -> H2 SO3 (aq) ( sulphuric(IV)acid turn litmus red)
13 4.Dry litmus papers and wet/damp/moist litmus papers were put in a gas jar containing sulphur(IV) oxide gas. State and explain the observations made. Observations (i)Dry Blue litmus paper remains blue. Dry red litmus paper remains red. (ii) Wet/damp/moist blue litmus paper turns red. Moist/damp/wet red litmus paper remains red. Both litmus papers are then bleached /decolorized. Explanation Dry sulphur(IV) oxide gas is a molecular compound that does not dissociate/ionize to release H+(aq)ions and thus has no effect on dry blue/red litmus papers. Wet/damp/moist litmus papers contain water that dissolves /react with dry sulphur(IV) oxide gas to form a solution of weak sulphuric(IV)acid (H2 SO3 (aq)). Weak sulphuric(IV)acid(H2 SO3 (aq)) dissociates /ionizes into free H+(aq)ions: H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) The free H+(aq)ions are responsible for turning blue litmus paper turns red showing the gas is acidic. The SO32- (aq) ions in wet/damp/moist sulphur(IV) oxide gas is responsible for many reactions of the gas. It is easily/readily oxidized to sulphate(VI) SO42- (aq) ions making sulphur(IV) oxide gas act as a reducing agent as in the following examples: (a)Bleaching agent Wet/damp/moist coloured flowers/litmus papers are bleached/decolorized when put in sulphur(IV) oxide gas.
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Weak sulphuric(IV)acid(H2 SO3 (aq)) dissociates /ionizes into free H+(aq)ions: H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) The free H+(aq)ions are responsible for turning blue litmus paper turns red showing the gas is acidic. The SO32- (aq) ions in wet/damp/moist sulphur(IV) oxide gas is responsible for many reactions of the gas. It is easily/readily oxidized to sulphate(VI) SO42- (aq) ions making sulphur(IV) oxide gas act as a reducing agent as in the following examples: (a)Bleaching agent Wet/damp/moist coloured flowers/litmus papers are bleached/decolorized when put in sulphur(IV) oxide gas. This is because sulphur(IV) oxide removes atomic oxygen from the coloured dye/ material to form sulphuric(VI)acid. Chemical equations (i)Formation of sulphuric(IV)acid SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Decolorization/bleaching of the dye/removal of atomic oxygen. Method I. H2 SO3 (aq) + (dye + O) -> H2 SO4 (aq) + dye (coloured) (colourless) Method II. H2 SO3 (aq) + (dye) -> H2 SO4 (aq) + (dye - O) (coloured) (colourless)
14 Sulphur(IV) oxide gas therefore bleaches by reduction /removing oxygen from a dye unlike chlorine that bleaches by oxidation /adding oxygen. The bleaching by removing oxygen from Sulphur(IV) oxide gas is temporary. This is because the bleached dye regains the atomic oxygen from the atmosphere/air in presence of sunlight as catalyst thus regaining/restoring its original colour. e.g. Old newspapers turn brown on exposure to air on regaining the atomic oxygen. The bleaching through adding oxygen by chlorine gas is permanent. (b)Turns Orange acidified potassium dichromate(VI) to green Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium dichromate(VI) solution.
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Old newspapers turn brown on exposure to air on regaining the atomic oxygen. The bleaching through adding oxygen by chlorine gas is permanent. (b)Turns Orange acidified potassium dichromate(VI) to green Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium dichromate(VI) solution. or; (ii)Dip a filter paper soaked in acidified potassium dichromate(VI) into a gas jar containing Sulphur(IV) oxide gas. Observation: Orange acidified potassium dichromate(VI) turns to green. Explanation: Sulphur(IV) oxide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: (i)Reaction of Sulphur(IV) oxide gas with water SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Dissociation /ionization of Sulphuric(IV)acid. H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) (iii)Oxidation of SO32- (aq)and reduction of Cr2O72-(aq) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) This is a confirmatory test for the presence of Sulphur(IV) oxide gas. Hydrogen sulphide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow residue. (c)Decolorizes acidified potassium manganate(VII)
15 Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or; (ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Sulphur(IV) oxide gas.
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Hydrogen sulphide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow residue. (c)Decolorizes acidified potassium manganate(VII)
15 Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or; (ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Sulphur(IV) oxide gas. Observation: Purple acidified potassium manganate(VII) turns to colourless/ acidified potassium manganate(VII) is decolorized. Explanation: Sulphur(IV) oxide gas reduces acidified potassium manganate(VII) from purple MnO4- ions to green Mn2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: (i)Reaction of Sulphur(IV) oxide gas with water SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Dissociation /ionization of Sulphuric(IV)acid. H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) (iii)Oxidation of SO32- (aq)and reduction of MnO4- (aq) 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) This is another test for the presence of Sulphur(IV) oxide gas. Hydrogen sulphide also decolorizes acidified potassium manganate(VII) from purple MnO4- ions to colourless Mn2+ ions leaving a yellow residue. (d)Decolorizes bromine water Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing bromine water . or; (ii)Put three drops of bromine water into a gas jar containing Sulphur(IV) oxide gas. Swirl.
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(d)Decolorizes bromine water Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing bromine water . or; (ii)Put three drops of bromine water into a gas jar containing Sulphur(IV) oxide gas. Swirl. Observation: Yellow bromine water turns to colourless/ bromine water is decolorized. Explanation:
16 Sulphur(IV) oxide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) without leaving a residue itself oxidized from SO32- ions in sulphuric (IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: (i)Reaction of Sulphur(IV) oxide gas with water SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Dissociation /ionization of Sulphuric(IV)acid. H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) (iii)Oxidation of SO32- (aq)and reduction of MnO4- (aq) SO32-(aq) + Br2 (aq) + H2O(l) -> SO42-(aq) + 2HBr(aq) (yellow) (colourless) This can also be used as another test for the presence of Sulphur(IV) oxide gas. Hydrogen sulphide also decolorizes yellow bromine water to colourless leaving a yellow residue. (e)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+ Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or; (ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Sulphur(IV) oxide gas.Swirl. Observation: Yellow/brown Iron (III)chloride solution turns to green Explanation: Sulphur(IV) oxide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
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(e)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+ Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or; (ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Sulphur(IV) oxide gas.Swirl. Observation: Yellow/brown Iron (III)chloride solution turns to green Explanation: Sulphur(IV) oxide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: (i)Reaction of Sulphur(IV) oxide gas with water SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Dissociation /ionization of Sulphuric(IV)acid. H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) (iii)Oxidation of SO32- (aq)and reduction of Fe3+ (aq) SO32-(aq) + 2Fe3+ (aq) +3H2O(l) -> SO42-(aq) + 2Fe2+(aq) + 2H+(aq) (yellow) (green)
17 (f)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or; (ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing Sulphur(IV) oxide gas. Swirl. Observation: Brown fumes of a gas evolved/produced. Explanation: Sulphur(IV) oxide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
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Swirl. Observation: Brown fumes of a gas evolved/produced. Explanation: Sulphur(IV) oxide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: SO2(g) + 2HNO3 (l) -> H2 SO4 (l) + NO2 (g) (brown fumes/gas) (g)Reduces Hydrogen peroxide to water Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide. Add four drops of Barium nitrate(V)or Barium chloride followed by five drops of 2M hydrochloric acid/ 2M nitric(V) acid. Observation: A white precipitate is formed that persist /remains on adding 2M hydrochloric acid/ 2M nitric(V) acid. Explanation: Sulphur(IV) oxide gas reduces 20 volume hydrogen peroxide and itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. When Ba2+ ions in Barium Nitrate(V) or Barium chloride solution is added, a white precipitate of insoluble Barium salts is formed showing the presence of of either SO32- ,SO42- ,CO32- ions. i.e. Chemical/ionic equation: SO32-(aq) + Ba2+ (aq) -> BaSO3(s) white precipitate SO42-(aq) + Ba2+ (aq) -> BaSO4(s) white precipitate CO32-(aq) + Ba2+ (aq) -> BaCO3(s) white precipitate
18 If nitric(V)/hydrochloric acid is added to the three suspected insoluble white precipitates above, the white precipitate: (i) persist/remains if SO42-(aq)ions (BaSO4(s)) is present. (ii)dissolves if SO32-(aq)ions (BaSO3(s)) and CO32-(aq)ions (BaCO3(s))is present. This is because: I.
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Chemical/ionic equation: SO32-(aq) + Ba2+ (aq) -> BaSO3(s) white precipitate SO42-(aq) + Ba2+ (aq) -> BaSO4(s) white precipitate CO32-(aq) + Ba2+ (aq) -> BaCO3(s) white precipitate
18 If nitric(V)/hydrochloric acid is added to the three suspected insoluble white precipitates above, the white precipitate: (i) persist/remains if SO42-(aq)ions (BaSO4(s)) is present. (ii)dissolves if SO32-(aq)ions (BaSO3(s)) and CO32-(aq)ions (BaCO3(s))is present. This is because: I. BaSO3(s) reacts with Nitric(V)/hydrochloric acid to produce acidic SO2 gas that turns Orange moist filter paper dipped in acidified Potassium dichromate to green. Chemical equation BaSO3(s) +2H+(aq) -> Ba2+ (aq) + SO2(g) + H2O(l) I. BaCO3(s) reacts with Nitric(V)/hydrochloric acid to produce acidic CO2 gas that forms a white precipitate when bubbled in lime water. Chemical equation BaCO3(s) +2H+(aq) -> Ba2+ (aq) + CO2(g) + H2O(l) 5.Sulphur(IV)oxide also act as an oxidizing agent as in the following examples. (a)Reduction by burning Magnesium Experiment Lower a burning Magnesium ribbon into agas jar containing Sulphur(IV)oxide gas Observation Magnesium ribbon continues to burn with difficulty. White ash and yellow powder/speck Explanation Sulphur(IV)oxide does not support burning/combustion. Magnesium burns to produce enough heat energy to decompose Sulphur(IV)oxide to sulphur and oxygen. The metal continues to burn on Oxygen forming white Magnesium oxide solid/ash. Yellow specks of sulphur residue form on the sides of reaction flask/gas jar.
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Magnesium burns to produce enough heat energy to decompose Sulphur(IV)oxide to sulphur and oxygen. The metal continues to burn on Oxygen forming white Magnesium oxide solid/ash. Yellow specks of sulphur residue form on the sides of reaction flask/gas jar. During the reaction, Sulphur(IV)oxide is reduced(oxidizing agent)while the metal is oxidized (reducing agent) Chemical equation SO2(g) + 2Mg(s) -> 2MgO(s) + S(s) (white ash/solid) (yellow speck/powder)
19 (b)Reduction by Hydrogen sulphide gas Experiment Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas Bubble hydrogen sulphide gas into the gas jar containing Sulphur(IV)oxide gas. Or Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas Invert a gas jar full of hydrogen sulphide gas over the gas jar containing Sulphur(IV)oxide gas. Swirl Observation Yellow powder/speck Explanation Sulphur(IV)oxide oxidizes hydrogen sulphide to yellow specks of sulphur residue and itself reduced to also sulphur that form on the sides of reaction flask/gas jar. A little moisture/water act as catalyst /speeds up the reaction. Chemical equation SO2(g) + 2H2S(g) -> 2H2O(l) + 3S(s) (yellow speck/powder) 6.Sulphur(IV)oxide has many industrial uses. State three. (i)In the contact process for the manufacture of Sulphuric(VI)acid (ii)As a bleaching agent of pulp and paper. (iii)As a fungicide to kill microbesβ (iv)As a preservative of jam, juices to prevent fermentation (ii) Sulphur(VI)oxide(SO3) (a) Occurrence Sulphur (VI)oxide is does not occur free in nature/atmosphere (b) Preparation In a Chemistry school laboratory Sulphur (VI)oxide may prepared from: Method 1;Catalytic oxidation of sulphur(IV)oxide gas. Sulphur(IV)oxide gas and oxygen mixture are first dried by being passed through Concentrated Sulphuric(VI)acid .
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(i)In the contact process for the manufacture of Sulphuric(VI)acid (ii)As a bleaching agent of pulp and paper. (iii)As a fungicide to kill microbesβ (iv)As a preservative of jam, juices to prevent fermentation (ii) Sulphur(VI)oxide(SO3) (a) Occurrence Sulphur (VI)oxide is does not occur free in nature/atmosphere (b) Preparation In a Chemistry school laboratory Sulphur (VI)oxide may prepared from: Method 1;Catalytic oxidation of sulphur(IV)oxide gas. Sulphur(IV)oxide gas and oxygen mixture are first dried by being passed through Concentrated Sulphuric(VI)acid . The dry mixture is then passed through platinised asbestos to catalyse/speed up the combination to form Sulphur (VI)oxide gas. Sulphur (VI)oxide gas readily solidify as silky white needles if passed through a freezing mixture /ice cold water. 20 The solid fumes out on heating to a highly acidic poisonous gas. Chemical equation 2SO2(g) + O2(g) --platinised asbestos--> 2SO3 (g) Method 2; Heating Iron(II)sulphate(VI) heptahydrate When green hydrated Iron(II)sulphate(VI) heptahydrate crystals are heated in a boiling tube ,it loses the water of crystallization and colour changes from green to white. Chemical equation FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l) (green solid) (white solid) On further heating ,the white anhydrous Iron(II)sulphate(VI) solid decomposes to a mixture of Sulphur (VI)oxide and Sulphur (IV)oxide gas. Sulphur (VI) oxide readily / easily solidify as white silky needles when the mixture is passed through a freezing mixture/ice cold water. Iron(III)oxide is left as a brown residue/solid.
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Chemical equation FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l) (green solid) (white solid) On further heating ,the white anhydrous Iron(II)sulphate(VI) solid decomposes to a mixture of Sulphur (VI)oxide and Sulphur (IV)oxide gas. Sulphur (VI) oxide readily / easily solidify as white silky needles when the mixture is passed through a freezing mixture/ice cold water. Iron(III)oxide is left as a brown residue/solid. Chemical equation 2FeSO4 (s) -> Fe2O3(s) + SO2 (g) + SO3(g) (green solid) (brown solid) Caution On exposure to air Sulphur (VI)oxide gas produces highly corrosive poisonous fumes of concentrated sulphuric(VI)acid and thus its preparation in a school laboratory is very risky. (c) Uses of sulphur(VI)oxide One of the main uses of sulphur(VI)oxide gas is as an intermediate product in the contact process for industrial/manufacture/large scale/production of sulphuric(VI)acid. (iii) Sulphuric(VI)acid(H2SO4) (a) Occurrence Sulphuric (VI)acid(H2SO4) is one of the three mineral acids. There are three mineral acids; Nitric(V)acid Sulphuric(VI)acid Hydrochloric acid. 21 Mineral acids do not occur naturally but are prepared in a school laboratory and manufactured at industrial level. (b)The Contact process for industrial manufacture of H2SO4 . I. Raw materials The main raw materials for industrial preparation of Sulphuric(VI)acid include: (i)Sulphur from Fraschs process or from heating metal sulphide ore like Galena(PbS),Zinc blende(ZnS) (ii)Oxygen from fractional distillation of air (iii)Water from rivers/lakes II. Chemical processes The contact process involves four main chemical processes: (i)Production of Sulphur (IV)oxide As one of the raw materials, Sulphur (IV)oxide gas is got from the following sources; I. Burning/roasting sulphur in air.
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Raw materials The main raw materials for industrial preparation of Sulphuric(VI)acid include: (i)Sulphur from Fraschs process or from heating metal sulphide ore like Galena(PbS),Zinc blende(ZnS) (ii)Oxygen from fractional distillation of air (iii)Water from rivers/lakes II. Chemical processes The contact process involves four main chemical processes: (i)Production of Sulphur (IV)oxide As one of the raw materials, Sulphur (IV)oxide gas is got from the following sources; I. Burning/roasting sulphur in air. Sulphur from Fraschs process is roasted/burnt in air to form Sulphur (IV)oxide gas in the burners Chemical equation S(s) + O2(g) --> SO2 (g) II. Burning/roasting sulphide ores in air. Sulphur (IV)oxide gas is produced as a by product in extraction of some metals like: - Lead from Lead(II)sulphide/Galena,(PbS) - Zinc from zinc(II)sulphide/Zinc blende, (ZnS) - Copper from Copper iron sulphide/Copper pyrites, (CuFeS2) On roasting/burning, large amount /quantity of sulphur(IV)oxide is generated/produced. Chemical equation (i)2PbS (s) + 3O2 (g) -> 2PbO(s) + 2SO2 (g) (ii)2ZnS (s) + 3O2 (g) -> 2ZnO(s) + 2SO2 (g) (ii)2CuFeS2 (s) + 4O2 (g) -> 2FeO(s) + 3SO2 (g) + Cu2O(s) Sulphur(IV)oxide easily/readily liquefies and thus can be transported to a far distance safely. 22 (ii)Purification of Sulphur(IV)oxide Sulphur(IV)oxide gas contain dust particles and Arsenic(IV)oxide as impurities. These impurities βpoisonβ/impair the catalyst by adhering on/covering its surface. The impurities are removed by electrostatic precipitation method .
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22 (ii)Purification of Sulphur(IV)oxide Sulphur(IV)oxide gas contain dust particles and Arsenic(IV)oxide as impurities. These impurities βpoisonβ/impair the catalyst by adhering on/covering its surface. The impurities are removed by electrostatic precipitation method . In the contact process Platinum or Vanadium(V)oxide may be used. Vanadium(V)oxide is preferred because it is : (i) cheaper/less expensive (ii) less easily poisoned by impurities (iii)Catalytic conversion of Sulphur(IV)oxide to Sulphur(VI)oxide Pure and dry mixture of Sulphur (IV)oxide gas and Oxygen is heated to 450oC in a heat exchanger. The heated mixture is passed through long pipes coated with pellets of Vanadium (V)oxide catalyst. The close βcontactβ between the reacting gases and catalyst give the process its name. Vanadium (V)oxide catalyse the conversion/oxidation of Sulphur(IV)oxide to Sulphur(VI)oxide gas. Chemical equation 2SO2 (g) + O2(g) -- V2O5 --> 2SO2 (g) This reaction is exothermic (-βH) and the temperatures need to be maintained at around 450oC to ensure that: (i)reaction rate/time taken for the formation of Sulphur(VI)oxide is not too slow/long at lower temperatures below 450oC (ii) Sulphur(VI)oxide gas does not decompose back to Sulphur(IV)oxide gas and Oxygen gas at higher temperatures than 450oC. (iv)Conversion of Sulphur(VI)oxide of Sulphuric(VI)acid Sulphur(VI)oxide is the acid anhydride of concentrated Sulphuric(VI)acid. Sulphur(VI)oxide reacts with water to form thick mist of fine droplets of very/highly corrosive concentrated Sulphuric(VI)acid because the reaction is highly exothermic. To prevent this, Sulphur (VI)oxide is a passed up to meet downward flow of 98% Sulphuric(VI)acid in the absorption chamber/tower.
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(iv)Conversion of Sulphur(VI)oxide of Sulphuric(VI)acid Sulphur(VI)oxide is the acid anhydride of concentrated Sulphuric(VI)acid. Sulphur(VI)oxide reacts with water to form thick mist of fine droplets of very/highly corrosive concentrated Sulphuric(VI)acid because the reaction is highly exothermic. To prevent this, Sulphur (VI)oxide is a passed up to meet downward flow of 98% Sulphuric(VI)acid in the absorption chamber/tower. The reaction forms a very viscous oily liquid called Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid. Chemical equation H2SO4 (aq) + SO3 (g) -> H2S2O7 (l)
23 Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid is diluted carefully with distilled water to give concentrated sulphuric (VI) acid . Chemical equation H2S2O7 (l) + H2O (l) -> 2H2SO4 (l) The acid is stored ready for market/sale. III. Environmental effects of contact process Sulphur(VI)oxide and Sulphur(IV)oxide gases are atmospheric pollutants that form acid rain if they escape to the atmosphere. In the Contact process, about 2% of these gases do not form sulphuric (VI) acid. The following precautions prevent/minimize pollution from Contact process: (i)recycling back any unreacted Sulphur(IV)oxide gas back to the heat exchangers. (ii)dissolving Sulphur(VI)oxide gas in concentrated sulphuric (VI) acid instead of water. This prevents the formation of fine droplets of the corrosive/ toxic/poisonous fumes of concentrated sulphuric (VI) acid. (iii)scrubbing-This involves passing the exhaust gases through very tall chimneys lined with quicklime/calcium hydroxide solid. This reacts with Sulphur (VI)oxide gas forming harmless calcium(II)sulphate (IV) /CaSO3 Chemical equation Ca(OH)2 (aq) + SO2(g) --> CaSO3 (aq) + H2O (g) IV.
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This prevents the formation of fine droplets of the corrosive/ toxic/poisonous fumes of concentrated sulphuric (VI) acid. (iii)scrubbing-This involves passing the exhaust gases through very tall chimneys lined with quicklime/calcium hydroxide solid. This reacts with Sulphur (VI)oxide gas forming harmless calcium(II)sulphate (IV) /CaSO3 Chemical equation Ca(OH)2 (aq) + SO2(g) --> CaSO3 (aq) + H2O (g) IV. Uses of Sulphuric(VI)acid Sulphuric (VI) acid is used: (i) in making dyes and paint (ii)as acid in Lead-acid accumulator/battery (iii) for making soapless detergents (iv) for making sulphate agricultural fertilizers VI. Sketch chart diagram showing the Contact process
24 (c) Properties of Concentrated sulphuric(VI)acid (i)Concentrated sulphuric(VI)acid is a colourless oily liquid with a density of 1.84gcm-3.It has a boiling point of 338oC. (ii) Concentrated sulphuric(VI)acid is very soluble in water. The solubility /dissolution of the acid very highly exothermic. The concentrated acid should thus be diluted slowly in excess water. Water should never be added to the acid because the hot acid scatters highly corrosive fumes out of the container. (iii) Concentrated sulphuric (VI)acid is a covalent compound. It has no free H+ ions. Free H+ ions are responsible for turning the blue litmus paper red. Concentrated sulphuric (VI) acid thus do not change the blue litmus paper red. (iv) Concentrated sulphuric (VI)acid is hygroscopic. It absorbs water from the atmosphere and do not form a solution. This makes concentrated sulphuric (VI) acid very suitable as drying agent during preparation of gases. (v)The following are some chemical properties of concentrated sulphuric (VI) acid: I. As a dehydrating agent Experiment I; Put about four spatula end fulls of brown sugar and glucose in separate 10cm3 beaker. Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Allow to stand for about 10 minutes.
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(v)The following are some chemical properties of concentrated sulphuric (VI) acid: I. As a dehydrating agent Experiment I; Put about four spatula end fulls of brown sugar and glucose in separate 10cm3 beaker. Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Allow to stand for about 10 minutes. Observation; Colour( in brown sugar )change from brown to black. Colour (in glucose) change from white to black. 10cm3 beaker becomes very hot. Explanation
25 Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes chemically and physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from compounds. When added to sugar /glucose a vigorous reaction that is highly exothermic take place. The sugar/glucose is charred to black mass of carbon because the acid dehydrates the sugar/glucose leaving carbon. Caution This reaction is highly exothermic that start slowly but produce fine particles of carbon that if inhaled cause quick suffocation by blocking the lung villi. Chemical equation Glucose: C6H12O6(s) --conc.H2SO4--> 6C (s) + 6H2O(l) (white) (black) Sugar: C12H22O11(s) --conc.H2SO4--> 12C (s) +11H2O(l) (brown) (black) Experiment II; Put about two spatula end full of hydrated copper(II)sulphate(VI)crystals in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Warm . Observation; Colour change from blue to white. Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent.It removes physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from hydrated compounds. The acid dehydrates blue copper(II)sulphate to white anhydrous copper(II)sulphate . Chemical equation CuSO4.5H2O(s) --conc.H2SO4--> CuSO4 (s) + 5H2O(l) (blue) (white) Experiment III; Put about 4cm3 of absolute ethanol in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid.
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Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent.It removes physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from hydrated compounds. The acid dehydrates blue copper(II)sulphate to white anhydrous copper(II)sulphate . Chemical equation CuSO4.5H2O(s) --conc.H2SO4--> CuSO4 (s) + 5H2O(l) (blue) (white) Experiment III; Put about 4cm3 of absolute ethanol in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid. Place moist/damp/wet filter paper dipped in acidified potassium dichromate(VI)solution on the mouth of the boiling tube. Heat strongly. Caution: Absolute ethanol is highly flammable. Observation; Colourless gas produced. 26 Orange acidified potassium dichromate (VI) paper turns to green. Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes chemically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from compounds. The acid dehydrates ethanol to ethene gas at about 170oC. Ethene with =C=C= double bond turns orange acidified potassium dichromate (VI) paper turns to green. Chemical equation C2H5OH(l) --conc.H2SO4/170oC --> C2H4 (g) + H2O(l) NB: This reaction is used for the school laboratory preparation of ethene gas Experiment IV; Put about 4cm3 of methanoic acid in a boiling tube .Carefully add about 6 cm3 of concentrated sulphuric (VI) acid. Heat gently Caution: This should be done in a fume chamber/open Observation; Colourless gas produced. Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes chemically combined elements of water (Hydrogen and Oxygen in ratio 2:1)from compounds. The acid dehydrates methanoic acid to poisonous/toxic carbon(II)oxide gas.
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Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes chemically combined elements of water (Hydrogen and Oxygen in ratio 2:1)from compounds. The acid dehydrates methanoic acid to poisonous/toxic carbon(II)oxide gas. Chemical equation HCOOH(l) --conc.H2SO4 --> CO(g) + H2O(l) NB: This reaction is used for the school laboratory preparation of small amount carbon (II)oxide gas Experiment V; Put about 4cm3 of ethan-1,2-dioic/oxalic acid in a boiling tube .Carefully add about 6 cm3 of concentrated sulphuric (VI) acid. Pass any gaseous product through lime water.Heat gently Caution: This should be done in a fume chamber/open Observation; Colourless gas produced. Gas produced forms a white precipitate with lime water. Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent. 27 It removes chemically combined elements of water (Hydrogen and Oxygen in ratio 2:1)from compounds. The acid dehydrates ethan-1,2-dioic/oxalic acid to a mixture of poisonous/toxic carbon(II)oxide and carbon(IV)oxide gases. Chemical equation HOOCCOOH(l) --conc.H2SO4 --> CO(g) + CO2(g) + H2O(l) NB: This reaction is also used for the school laboratory preparation of small amount carbon (II) oxide gas. Carbon (IV) oxide gas is removed by passing the mixture through concentrated sodium/potassium hydroxide solution. II. As an Oxidizing agent Experiment I Put about 2cm3 of Concentrated sulphuric (VI) acid into three separate boiling tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution on the mouth of the boiling tube. Put about 0.5g of Copper turnings, Zinc granule and Iron filings to each boiling tube separately. Observation; Effervescence/fizzing/bubbles Blue solution formed with copper, Green solution formed with Iron Colourless solution formed with Zinc Colourless gas produced that has a pungent irritating choking smell. Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.
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Put about 0.5g of Copper turnings, Zinc granule and Iron filings to each boiling tube separately. Observation; Effervescence/fizzing/bubbles Blue solution formed with copper, Green solution formed with Iron Colourless solution formed with Zinc Colourless gas produced that has a pungent irritating choking smell. Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green. Explanation Concentrated sulphuric (VI) acid is strong oxidizing agent. It oxidizes metals to metallic sulphate(VI) salts and itself reduced to sulphur(IV)oxide gas. Sulphur (IV) oxide gas turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green. CuSO4(aq) is a blue solution. ZnSO4(aq) is a colourless solution. FeSO4(aq) is a green solution. Chemical equation Cu(s) + 2H2SO4(aq) --> CuSO4(aq) + SO2(g) + 2H2O(l) Zn(s) + 2H2SO4(aq) --> ZnSO4(aq) + SO2(g) + 2H2O(l) Fe(s) + 2H2SO4(aq) --> FeSO4(aq) + SO2(g) + 2H2O(l) Experiment II
28 Put about 2cm3 of Concentrated sulphuric (VI) acid into two separate boiling tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution on the mouth of the boiling tube. Put about 0.5g of powdered charcoal and sulphur powder to each boiling tube separately. Warm. Observation; Black solid charcoal dissolves/decrease Yellow solid sulphur dissolves/decrease Colourless gas produced that has a pungent irritating choking smell. Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green. Explanation Concentrated sulphuric (VI) acid is strong oxidizing agent. It oxidizes nonmetals to non metallic oxides and itself reduced to sulphur(IV)oxide gas.
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Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green. Explanation Concentrated sulphuric (VI) acid is strong oxidizing agent. It oxidizes nonmetals to non metallic oxides and itself reduced to sulphur(IV)oxide gas. Sulphur (IV) oxide gas turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green. Charcoal is oxidized to carbon(IV)oxide. Sulphur is oxidized to Sulphur(IV)oxide . Chemical equation C(s) + 2H2SO4(aq) --> CO2(aq) + 2SO2(g) + 2H2O(l) S(s) + 2H2SO4(aq) --> 3SO2(g) + 2H2O(l) III. As the least volatile acid Study the table below showing a comparison in boiling points of the three mineral acids Mineral acid Relative molecula mass Boiling point(oC) Hydrochloric acid(HCl) 36.5 35.0 Nitric(V)acid(HNO3) 63.0 83.0 Sulphuric(VI)acid(H2SO4) 98.0 333 1.Which is the least volatile acid? Explain Sulphuric(VI)acid(H2SO4) because it has the largest molecule and joined by Hydrogen bonds making it to have the highest boiling point/least volatile. 2. Using chemical equations, explain how sulphuric(VI)acid displaces the less volatile mineral acids. (i)Chemical equation KNO3(s) + H2SO4(aq) --> KHSO4(l) + HNO3(g) NaNO3(s) + H2SO4(aq) --> NaHSO4(l) + HNO3(g)
29 This reaction is used in the school laboratory preparation of Nitric(V) acid (HNO3). (ii)Chemical equation KCl(s) + H2SO4(aq) --> KHSO4(s) + HCl(g) NaCl(s) + H2SO4(aq) --> NaHSO4(s) + HCl(g) This reaction is used in the school laboratory preparation of Hydrochloric acid (HCl). (d) Properties of dilute sulphuric(VI)acid.
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(i)Chemical equation KNO3(s) + H2SO4(aq) --> KHSO4(l) + HNO3(g) NaNO3(s) + H2SO4(aq) --> NaHSO4(l) + HNO3(g)
29 This reaction is used in the school laboratory preparation of Nitric(V) acid (HNO3). (ii)Chemical equation KCl(s) + H2SO4(aq) --> KHSO4(s) + HCl(g) NaCl(s) + H2SO4(aq) --> NaHSO4(s) + HCl(g) This reaction is used in the school laboratory preparation of Hydrochloric acid (HCl). (d) Properties of dilute sulphuric(VI)acid. Dilute sulphuric(VI)acid is made when about 10cm3 of concentrated sulphuric (VI) acid is carefully added to about 90cm3 of distilled water. Diluting concentrated sulphuric (VI) acid should be done carefully because the reaction is highly exothermic. Diluting concentrated sulphuric (VI) acid decreases the number of moles present in a given volume of solution which makes the acid less corrosive. On diluting concentrated sulphuric(VI) acid, water ionizes /dissociates the acid fully/wholly into two(dibasic)free H+(aq) and SO42-(aq)ions: H2SO4 (aq) -> 2H+(aq) + SO42-(aq) The presence of free H+(aq)ions is responsible for ; (i)turn litmus red because of the presence of free H+(aq)ions (ii)have pH 1/2/3 because of the presence of many free H+(aq)ions hence a strongly acidic solution. (iii)Reaction with metals Experiment: Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean test tubes. Add about 0.1g of Magnesium ribbon to one test tube. Cover the mixture with a finger as stopper. Introduce a burning splint on top of the finger and release the finger βstopperβ. Repeat by adding Zinc, Copper and Iron instead of the Magnesium ribbon.
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Cover the mixture with a finger as stopper. Introduce a burning splint on top of the finger and release the finger βstopperβ. Repeat by adding Zinc, Copper and Iron instead of the Magnesium ribbon. Observation: No effervescence/ bubbles/ fizzing with copper Effervescence/ bubbles/ fizzing with Iron ,Zinc and Magnesium Colourless gas produced that extinguishes burning splint with a βpopβ sound. Colourless solution formed with Zinc and Magnesium. Green solution formed with Iron Explanation: When a metal higher than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI)acid,
30 effervescence/ bubbling/ fizzing takes place with evolution of Hydrogen gas. Impure hydrogen gas extinguishes burning splint with a βpopβ sound. A sulphate (VI) salts is formed. Iron, Zinc and Magnesium are higher than hydrogen in the reactivity/electrochemical series. They form Iron (II)sulphate(VI), Magnesium sulphate(VI) and Zinc sulphate(VI). . When a metal lower than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI)acid, there is no effervescence/ bubbling/ fizzing that take place. Copper thus do not react with dilute sulphuric(VI)acid. Chemical/ionic equation Mg(s) + H2SO4(aq) --> MgSO4(aq) + H2(g) Mg(s) + 2H+(aq) --> Mg2+ (aq) + H2(g) Zn(s) + H2SO4(aq) --> ZnSO4(aq) + H2(g) Zn(s) + 2H+(aq) --> Zn2+ (aq) + H2(g) Fe(s) + H2SO4(aq) --> FeSO4(aq) + H2(g) Fe(s) + H+(aq) --> Fe2+ (aq) + H2(g) NB:(i) Calcium,Lead and Barium forms insoluble sulphate(VI)salts that cover/coat the unreacted metals.
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When a metal lower than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI)acid, there is no effervescence/ bubbling/ fizzing that take place. Copper thus do not react with dilute sulphuric(VI)acid. Chemical/ionic equation Mg(s) + H2SO4(aq) --> MgSO4(aq) + H2(g) Mg(s) + 2H+(aq) --> Mg2+ (aq) + H2(g) Zn(s) + H2SO4(aq) --> ZnSO4(aq) + H2(g) Zn(s) + 2H+(aq) --> Zn2+ (aq) + H2(g) Fe(s) + H2SO4(aq) --> FeSO4(aq) + H2(g) Fe(s) + H+(aq) --> Fe2+ (aq) + H2(g) NB:(i) Calcium,Lead and Barium forms insoluble sulphate(VI)salts that cover/coat the unreacted metals. (ii)Sodium and Potassium react explosively with dilute sulphuric(VI)acid (iv)Reaction with metal carbonates and hydrogen carbonates Experiment: Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean boiling tubes. Add about 0.1g of sodium carbonate to one boiling tube. Introduce a burning splint on top of the boiling tube. Repeat by adding Zinc carbonate, Copper (II)carbonate and Iron(II)Carbonate in place of the sodium hydrogen carbonate. Observation: Effervescence/ bubbles/ fizzing. Colourless gas produced that extinguishes burning splint. Colourless solution formed with Zinc carbonate, sodium hydrogen carbonate and sodium carbonate. Green solution formed with Iron(II)Carbonate Blue solution formed with Copper(II)Carbonate Explanation: When a metal carbonate or a hydrogen carbonates is put in a test tube containing dilute sulphuric(VI)acid, effervescence/ bubbling/ fizzing takes place with evolution of carbon(IV)oxide gas. carbon(IV)oxide gas
31 extinguishes a burning splint and forms a white precipitate when bubbled in lime water. A sulphate (VI) salts is formed.
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Green solution formed with Iron(II)Carbonate Blue solution formed with Copper(II)Carbonate Explanation: When a metal carbonate or a hydrogen carbonates is put in a test tube containing dilute sulphuric(VI)acid, effervescence/ bubbling/ fizzing takes place with evolution of carbon(IV)oxide gas. carbon(IV)oxide gas
31 extinguishes a burning splint and forms a white precipitate when bubbled in lime water. A sulphate (VI) salts is formed. Chemical/ionic equation ZnCO3(s) + H2SO4(aq) --> ZnSO4(aq) + H2O(l) + CO2(g) ZnCO3(s) + 2H+(aq) --> Zn2+ (aq) + H2O(l) + CO2(g) CuCO3(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l) + CO2(g) CuCO3(s) + 2H+(aq) --> Cu2+ (aq) + H2O(l) + CO2(g) FeCO3(s) + H2SO4(aq) --> FeSO4(aq) + H2O(l) + CO2(g) FeCO3(s) + 2H+(aq) --> Fe2+ (aq) + H2O(l) + CO2(g) 2NaHCO3(s) + H2SO4(aq) --> Na2SO4(aq) + 2H2O(l) + 2CO2(g) NaHCO3(s) + H+(aq) --> Na+ (aq) + H2O(l) + CO2(g) Na2CO3(s) + H2SO4(aq) --> Na2SO4(aq) + H2O(l) + CO2(g) NaHCO3(s) + H+(aq) --> Na+ (aq) + H2O(l) + CO2(g) (NH4)2CO3(s) + H2SO4(aq) --> (NH4)2SO4 (aq) + H2O(l) + CO2(g) (NH4)2CO3 (s) + H+(aq) --> NH4+ (aq) + H2O(l) + CO2(g) 2NH4HCO3(aq) + H2SO4(aq) --> (NH4)2SO4 (aq) + H2O(l) + CO2(g) NH4HCO3(aq) + H+(aq) --> NH4+ (aq) + H2O(l) + CO2(g) NB: Calcium, Lead and Barium carbonates forms insoluble sulphate(VI)salts that cover/coat the unreacted metals.
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(v)Neutralization-reaction of metal oxides and alkalis/bases Experiment I: Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean boiling tubes. Add about 0.1g of copper(II)oxide to one boiling tube. Stir. Repeat by adding Zinc oxide, calcium carbonate and Sodium (II)Oxide in place of the Copper(II)Oxide. Observation: Blue solution formed with Copper(II)Oxide Colourless solution formed with other oxides Explanation: When a metal oxide is put in a test tube containing dilute sulphuric(VI)acid, the oxide dissolves forming a sulphate (VI) salt. Chemical/ionic equation ZnO(s) + H2SO4(aq) --> ZnSO4(aq) + H2O(l) ZnO(s) + 2H+(aq) --> Zn2+ (aq) + H2O(l)
32 CuO(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l) CuO(s) + 2H+(aq) --> Cu2+ (aq) + H2O(l) MgO(s) + H2SO4(aq) --> MgSO4(aq) + H2O(l) MgO(s) + 2H+(aq) --> Mg2+ (aq) + H2O(l) Na2O(s) + H2SO4(aq) --> Na2SO4(aq) + H2O(l) Na2O(s) + 2H+(aq) --> 2Na+ (aq) + H2O(l) K2CO3(s) + H2SO4(aq) --> K2SO4(aq) + H2O(l) K2O(s) + H+(aq) --> 2K+ (aq) + H2O(l) NB: Calcium, Lead and Barium oxides forms insoluble sulphate(VI)salts that cover/coat the unreacted metals oxides. Experiment II: Fill a burette wuth 0.1M dilute sulphuric(VI)acid. Pipette 20.0cm3 of 0.1Msodium hydroxide solution into a 250cm3 conical flask.
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Chemical/ionic equation ZnO(s) + H2SO4(aq) --> ZnSO4(aq) + H2O(l) ZnO(s) + 2H+(aq) --> Zn2+ (aq) + H2O(l)
32 CuO(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l) CuO(s) + 2H+(aq) --> Cu2+ (aq) + H2O(l) MgO(s) + H2SO4(aq) --> MgSO4(aq) + H2O(l) MgO(s) + 2H+(aq) --> Mg2+ (aq) + H2O(l) Na2O(s) + H2SO4(aq) --> Na2SO4(aq) + H2O(l) Na2O(s) + 2H+(aq) --> 2Na+ (aq) + H2O(l) K2CO3(s) + H2SO4(aq) --> K2SO4(aq) + H2O(l) K2O(s) + H+(aq) --> 2K+ (aq) + H2O(l) NB: Calcium, Lead and Barium oxides forms insoluble sulphate(VI)salts that cover/coat the unreacted metals oxides. Experiment II: Fill a burette wuth 0.1M dilute sulphuric(VI)acid. Pipette 20.0cm3 of 0.1Msodium hydroxide solution into a 250cm3 conical flask. Add three drops of phenolphthalein indicator.Titrate the acid to get a permanent colour change.Repeat with0.1M potassium hydroxide solution inplace of 0.1Msodium hydroxide solution Observation: Colour of phenolphthalein changes from pink to colourless at the end point. Explanation Like other (mineral) acids dilute sulphuric(VI)acid neutralizes bases/alkalis to a sulphate salt and water only.
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Pipette 20.0cm3 of 0.1Msodium hydroxide solution into a 250cm3 conical flask. Add three drops of phenolphthalein indicator.Titrate the acid to get a permanent colour change.Repeat with0.1M potassium hydroxide solution inplace of 0.1Msodium hydroxide solution Observation: Colour of phenolphthalein changes from pink to colourless at the end point. Explanation Like other (mineral) acids dilute sulphuric(VI)acid neutralizes bases/alkalis to a sulphate salt and water only. Colour of the indicator used changes when a slight excess of acid is added to the base at the end point Chemical equation: 2NaOH(aq) + H2SO4(aq) --> Na2SO4(aq) + H2O(l) OH-(s) + H+(aq) --> H2O(l) 2KOH(aq) + H2SO4(aq) --> K2SO4(aq) + H2O(l) OH-(s) + H+(aq) --> H2O(l) 2NH4OH(aq) + H2SO4(aq) --> (NH4)2SO4(aq) + H2O(l) OH-(s) + H+(aq) --> H2O(l)
33 (iv) Hydrogen sulphide(H2S) (a) Occurrence Hydrogen sulphide is found in volcanic areas as a gas or dissolved in water from geysers and hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya. It is present in rotten eggs and human excreta. (b) Preparation Hydrogen sulphide is prepared in a school laboratory by heating Iron (II) sulphide with dilute hydrochloric acid. (c) Properties of Hydrogen sulphide(Questions) 1. Write the equation for the reaction for the school laboratory preparation of Hydrogen sulphide. 34 Chemical equation: FeS (s) + 2HCl (aq) -> H2S (g) FeCl2 (aq) 2. State three physical properties unique to Hydrogen sulphide. Hydrogen sulphide is a colourless gas with characteristic pungent poisonous smell of rotten eggs. It is soluble in cold water but insoluble in warm water.
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State three physical properties unique to Hydrogen sulphide. Hydrogen sulphide is a colourless gas with characteristic pungent poisonous smell of rotten eggs. It is soluble in cold water but insoluble in warm water. It is denser than water and turns blue litmus paper red. 3. Hydrogen sulphide exist as a dibasic acid when dissolved in water. Using a chemical equation show how it ionizes in aqueous state. H2S(aq) -> H+(aq) + HS-(aq) H2S(aq) -> 2H+(aq) + S2- (aq) Hydrogen sulphide therefore can form both normal and acid salts e.g Sodium hydrogen sulphide and sodium sulphide both exist 4. State and explain one gaseous impurity likely to be present in the gas jar containing hydrogen sulphide above. Hydrogen/ H2 Iron(II)sulphide contains Iron as impurity .The iron will react with dilute hydrochloric acid to form iron(II)chloride and produce hydrogen gas that mixes with hydrogen sulphide gas. 5. State and explain the observations made when a filter paper dipped in Lead(II) ethanoate /Lead (II) nitrate(V) solution is put in a gas jar containing hydrogen sulphide gas. Observations Moist Lead(II) ethanoate /Lead (II) nitrate(V) paper turns black. Explanation When hydrogen sulphide is bubbled in a metallic salt solution, a metallic sulphide is formed. All sulphides are insoluble black salts except sodium sulphide, potassium sulphide and ammonium sulphides. Hydrogen sulphide gas blackens moist Lead (II) ethanoate /Lead (II) nitrate(V) paper . The gas reacts with Pb2+ in the paper to form black Lead(II)sulphide. This is the chemical test for the presence of H2S other than the physical smell of rotten eggs.
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Hydrogen sulphide gas blackens moist Lead (II) ethanoate /Lead (II) nitrate(V) paper . The gas reacts with Pb2+ in the paper to form black Lead(II)sulphide. This is the chemical test for the presence of H2S other than the physical smell of rotten eggs. Chemical equations Pb2+(aq) + H2S -> PbS + 2H+(aq) (black) Fe2+(aq) + H2S -> FeS + 2H+(aq) (black) Zn2+(aq) + H2S -> ZnS + 2H+(aq) (black) Cu2+(aq) + H2S -> CuS + 2H+(aq) (black)
35 2Cu+(aq) + H2S -> Cu2S + 2H+(aq) (black) 6. Dry hydrogen sulphide was ignited as below. (i) State the observations made in flame A Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV)oxide gas and water. Chemical equation: 2H2S(g) + 3O2(g) -> 2H2O(l) + 2SO2(g) Hydrogen sulphide burns in limited air with a blue flame to form sulphur solid and water. Chemical equation: 2H2S(g) + O2(g) -> 2H2O(l) + 2S(s) 7. Hydrogen sulphide is a strong reducing agent that is oxidized to yellow solid sulphur as precipitate. The following experiments illustrate the reducing properties of Hydrogen sulphide. (a)Turns Orange acidified potassium dichromate(VI) to green Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing acidified potassium dichromate (VI) solution. or; (ii)Dip a filter paper soaked in acidified potassium dichromate (VI) into a gas jar containing Hydrogen sulphide gas. Observation: Orange acidified potassium dichromate (VI) turns to green. Yellow solid residue. Explanation: Hydrogen sulphide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow solid residue as itself is oxidized to sulphur.
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Observation: Orange acidified potassium dichromate (VI) turns to green. Yellow solid residue. Explanation: Hydrogen sulphide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow solid residue as itself is oxidized to sulphur. Chemical/ionic equation: 4H2S(aq) + Cr2O72-(aq) +6H+(aq) -> 4S(aq) + 2Cr3+(aq) + 7H2O(l) This test is used for differentiating Hydrogen sulphide and sulphur (IV)oxide gas. Flame A Dry Hydrogen sulphide gas
36 Sulphur(IV)oxide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions without leaving a yellow residue. (b)Decolorizes acidified potassium manganate(VII) Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or; (ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Hydrogen Sulphide gas. Observation: Purple acidified potassium manganate(VII) turns to colourless/ acidified potassium manganate(VII) is decolorized. Yellow solid residue. Explanation: Hydrogen sulphide gas reduces acidified potassium manganate(VII) from purple MnO4- ions to green Mn2+ ions leaving a residue as the gas itself is oxidized to sulphur. Chemical/ionic equation: 5H2S(g) + 2MnO4- (aq) +6H+(aq) -> 5S (s) + 2Mn2+(aq) + 8H2O(l) (purple) (colourless) This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide gas. Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO4- ions to colourless Mn2+ ions leaving no yellow residue. (c)Decolorizes bromine water Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing bromine water .
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Chemical/ionic equation: 5H2S(g) + 2MnO4- (aq) +6H+(aq) -> 5S (s) + 2Mn2+(aq) + 8H2O(l) (purple) (colourless) This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide gas. Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO4- ions to colourless Mn2+ ions leaving no yellow residue. (c)Decolorizes bromine water Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing bromine water . or; (ii)Put three drops of bromine water into a gas jar containing Hydrogen sulphide gas. Swirl. Observation: Yellow bromine water turns to colourless/ bromine water is decolorized. Yellow solid residue Explanation: Hydrogen sulphide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) leaving a yellow residue as the gas itself is oxidized to sulphur. 37 Chemical/ionic equation: H2 S(g) + Br2 (aq) -> S (s) + 2HBr(aq) (yellow solution) (yellow solid) (colourless) This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide gas. Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO4- ions to colourless Mn2+ ions leaving no yellow residue. (d)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+ Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or; (ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Hydrogen sulphide gas. Swirl. Observation: Yellow/brown Iron (III)chloride solution turns to green. Yellow solid solid Explanation: Hydrogen sulphide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions leaving a yellow residue.The gas is itself oxidized to sulphur.
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Swirl. Observation: Yellow/brown Iron (III)chloride solution turns to green. Yellow solid solid Explanation: Hydrogen sulphide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions leaving a yellow residue.The gas is itself oxidized to sulphur. Chemical/ionic equation: H2S(aq) + 2Fe3+ (aq) -> S (s) + Fe2+(aq) + 2H+(aq) (yellow solution) (yellow residue) (green) (e)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or; (ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing Hydrogen sulphide gas. Swirl. Observation: Brown fumes of a gas evolved/produced. Yellow solid residue Explanation: Hydrogen sulphide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized to yellow sulphur. Chemical/ionic equation: H2S(g) + 2HNO3 (l) -> 2H2O(l) + S (s) + 2NO2 (g)
38 (yellow residue) (brown fumes) (f)Reduces sulphuric(VI)acid to Sulphur Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated sulphuric(VI)acid. or; (ii)Place about 3cm3 of concentrated sulphuric (VI) acid into a gas jar containing Hydrogen sulphide gas. Swirl. Observation: Yellow solid residue Explanation: Hydrogen sulphide gas reduces concentrated sulphuric(VI)acid to yellow sulphur. Chemical/ionic equation: 3H2S(g) + H2SO4 (l) -> 4H2O(l) + 4S (s) (yellow residue) (g)Reduces Hydrogen peroxide to water Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide.
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Swirl. Observation: Yellow solid residue Explanation: Hydrogen sulphide gas reduces concentrated sulphuric(VI)acid to yellow sulphur. Chemical/ionic equation: 3H2S(g) + H2SO4 (l) -> 4H2O(l) + 4S (s) (yellow residue) (g)Reduces Hydrogen peroxide to water Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide. Observation: Yellow solid residue Explanation: Hydrogen sulphide gas reduces 20 volume hydrogen peroxide to water and itself oxidized to yellow sulphur Chemical/ionic equation: H2S(g) + H2O2 (l) -> 2H2O(l) + S (s) (yellow residue) 8.Name the salt formed when: (i)equal volumes of equimolar hydrogen sulphide neutralizes sodium hydroxide solution: Sodium hydrogen sulphide Chemical/ionic equation: H2S(g) + NaOH (l) -> H2O(l) + NaHS (aq) (ii) hydrogen sulphide neutralizes excess concentrated sodium hydroxide solution: Sodium sulphide Chemical/ionic equation:
39 H2S(g) + 2NaOH (l) -> 2H2O(l) + Na2S (aq) Practice Hydrogen sulphide gas was bubbled into a solution of metallic nitrate(V)salts as in the flow chart below (a)Name the black solid Copper(II)sulphide (b)Identify the cation responsible for the formation of: I. Blue solution Cu2+(aq) II. Green solution Fe2+(aq) III. Brown solution Fe3+(aq) (c)Using acidified potassium dichromate(VI) describe how you would differentiate between sulphur(IV)Oxide and hydrogen sulphide -Bubble the gases in separate test tubes containing acidified Potassium dichromate(VI) solution. -Both changes the Orange colour of acidified Potassium dichromate(VI) solution to green.
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Green solution Fe2+(aq) III. Brown solution Fe3+(aq) (c)Using acidified potassium dichromate(VI) describe how you would differentiate between sulphur(IV)Oxide and hydrogen sulphide -Bubble the gases in separate test tubes containing acidified Potassium dichromate(VI) solution. -Both changes the Orange colour of acidified Potassium dichromate(VI) solution to green. -Yellow solid residue/deposit is formed with Hydrogen sulphide Chemical/ionic equation: 4H2S(aq) + Cr2O72-(aq) +6H+(aq) -> 4S(aq) + 2Cr3+(aq) + 7H2O(l) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (d)State and explain the observations made if a burning splint is introduced at the mouth of a hydrogen sulphide generator. ObservationGas continues burning with a blue flame Explanation: Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV)oxide gas and water. Chemical equation: 2H2S(g)+ 3O2(g) -> 2H2O(l) + 2SO2 (g) Hydrogen sulphide Blue solution Brown solution Black solid Green solution
40 (v)Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts 1. Sulphate (VI) (SO42-) salts are normal and acid salts derived from Sulphuric (VI)acid H2SO4. 2. Sulphate(IV) (SO32-) salts are normal and acid salts derived from Sulphuric (IV)acid H2SO3. 3. Sulphuric (VI)acid H2SO4 is formed when sulphur(VI)oxide gas is bubbled in water. The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (VI) (SO42-) and hydrogen sulphate (VI) (HSO4-) salts. i.e.
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The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (VI) (SO42-) and hydrogen sulphate (VI) (HSO4-) salts. i.e. H2SO4 (aq) -> 2H+(aq) + SO42-(aq) H2SO4 (aq) -> H+(aq) + HSO4 -(aq) All Sulphate (VI) (SO42-) salts dissolve in water/are soluble except Calcium (II) sulphate (VI) (CaSO4), Barium (II) sulphate (VI) (BaSO4) and Lead (II) sulphate (VI) (PbSO4) All Hydrogen sulphate (VI) (HSO3-) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (VI) (NaHSO4), Potassium (I) hydrogen sulphate (VI) (KHSO4) and Ammonium hydrogen sulphate (VI) (NH4HSO4) exist also as solids. Other Hydrogen sulphate (VI) (HSO4-) salts do not exist except those of Calcium (II) hydrogen sulphate (VI) (Ca (HSO4)2) and Magnesium (II) hydrogen sulphate (VI) (Mg (HSO4)2). 4. Sulphuric (IV)acid H2SO3 is formed when sulphur(IV)oxide gas is bubbled in water. The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (IV) (SO32-) and hydrogen sulphate (VI) (HSO4-) salts. i.e. H2SO3 (aq) -> 2H+(aq) + SO32-(aq)
41 H2SO3 (aq) -> H+(aq) + HSO3 -(aq) All Sulphate (IV) (SO32-) salts dissolve in water/are soluble except Calcium (II) sulphate (IV) (CaSO3), Barium (II) sulphate (IV) (BaSO3) and Lead (II) sulphate (IV) (PbSO3) All Hydrogen sulphate (IV) (HSO3-) salts exist in solution/dissolved in water.
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It forms therefore the Sulphate (IV) (SO32-) and hydrogen sulphate (VI) (HSO4-) salts. i.e. H2SO3 (aq) -> 2H+(aq) + SO32-(aq)
41 H2SO3 (aq) -> H+(aq) + HSO3 -(aq) All Sulphate (IV) (SO32-) salts dissolve in water/are soluble except Calcium (II) sulphate (IV) (CaSO3), Barium (II) sulphate (IV) (BaSO3) and Lead (II) sulphate (IV) (PbSO3) All Hydrogen sulphate (IV) (HSO3-) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (IV) (NaHSO3), Potassium (I) hydrogen sulphate (IV) (KHSO3) and Ammonium hydrogen sulphate (IV) (NH4HSO3) exist also as solids. Other Hydrogen sulphate (IV) (HSO3-) salts do not exist except those of Calcium (II) hydrogen sulphate (IV) (Ca (HSO3)2) and Magnesium (II) hydrogen sulphate (IV) (Mg (HSO3)2). 5.The following experiments show the effect of heat on sulphate(VI) (SO42-)and sulphate(IV) (SO32-) salts: Experiment: In a clean dry test tube place separately about 1.0g of : Zinc(II)sulphate (VI), Iron(II)sulphate(VI), Copper(II)sulphate(VI),Sodium (I) sulphate (VI), Sodium (I) sulphate (IV).Heat gently then strongly. Test any gases produced using litmus papers. Observations: -Colourless droplets of liquid forms on the cooler parts of the test tube in all cases. -White solid residue is left in case of Zinc (II)sulphate(VI),Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV).
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Test any gases produced using litmus papers. Observations: -Colourless droplets of liquid forms on the cooler parts of the test tube in all cases. -White solid residue is left in case of Zinc (II)sulphate(VI),Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV). -Colour changes from green to brown /yellow in case of Iron (II)sulphate(VI) -Colour changes from blue to white then black in case of Copper (II) sulphate (VI) -Blue litmus paper remain and blue and red litmus paper remain red in case of Zinc(II)sulphate(VI), Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV) -Blue litmus paper turns red and red litmus paper remain red in case of Iron (II)sulphate(VI) and Copper (II) sulphate (VI). Explanation (i)All Sulphate (VI) (SO42-) salts exist as hydrated salts with water of crystallization that condenses and collects on cooler parts of test tube as a colourless liquid on gentle heating. e.g. K2SO4.10H2O(s) -> K2SO4(s) + 10H2O(l) Na2SO4.10H2O(s) -> Na2SO4(s) + 10H2O(l) MgSO4.7H2O(s) -> MgSO4(s) + 7H2O(l) CaSO4.7H2O(s) -> CaSO4(s) + 7H2O(l) ZnSO4.7H2O(s) -> ZnSO4(s) + 7H2O(l) FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l) Al2(SO4)3.6H2O(s) -> Al2(SO4)3 (s) + 6H2O(l)
42 CuSO4.5H2O(s) -> CuSO4(s) + 5H2O(l) All Sulphate (VI) (SO42-) salts do not decompose on heating except Iron (II) sulphate (VI) and Copper (II) sulphate (VI).
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Explanation (i)All Sulphate (VI) (SO42-) salts exist as hydrated salts with water of crystallization that condenses and collects on cooler parts of test tube as a colourless liquid on gentle heating. e.g. K2SO4.10H2O(s) -> K2SO4(s) + 10H2O(l) Na2SO4.10H2O(s) -> Na2SO4(s) + 10H2O(l) MgSO4.7H2O(s) -> MgSO4(s) + 7H2O(l) CaSO4.7H2O(s) -> CaSO4(s) + 7H2O(l) ZnSO4.7H2O(s) -> ZnSO4(s) + 7H2O(l) FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l) Al2(SO4)3.6H2O(s) -> Al2(SO4)3 (s) + 6H2O(l)
42 CuSO4.5H2O(s) -> CuSO4(s) + 5H2O(l) All Sulphate (VI) (SO42-) salts do not decompose on heating except Iron (II) sulphate (VI) and Copper (II) sulphate (VI). (i)Iron (II) sulphate (VI) decomposes on strong heating to produce acidic sulphur (IV)oxide and sulphur(VI)oxide gases. Iron(III)oxide is formed as a brown /yellow residue. Chemical equation 2FeSO4 (s) -> Fe2O3(s) + SO2(g) + SO3(g) This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas. Sulphur (VI) oxide readily /easily solidifies as white silky needles when the mixture is passed through freezing mixture/ice cold water. Sulphur (IV) oxide does not. (ii) Copper(II)sulphate(VI) decomposes on strong heating to black copper (II) oxide and Sulphur (VI) oxide gas. Chemical equation 2CuSO4 (s) -> CuO(s) + SO3(g) This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas. 6.
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(ii) Copper(II)sulphate(VI) decomposes on strong heating to black copper (II) oxide and Sulphur (VI) oxide gas. Chemical equation 2CuSO4 (s) -> CuO(s) + SO3(g) This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas. 6. The following experiments show the test for the presence of sulphate (VI) (SO42-)and sulphate(IV) (SO32-) ions in a sample of a salt/compound: Experiments/Observations: (a)Using Lead(II)nitrate(V) I. To about 5cm3 of a salt solution in a test tube add four drops of Lead(II)nitrate(V)solution. Preserve. Observation Inference White precipitate/ppt SO42- , SO32- , CO32- , Cl- ions II. To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve. Observation 1 Observation Inference White precipitate/ppt persists SO42- , Cl- ions Observation 2 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.(a)To the preserved sample observation 1 in (II) above, Heat to boil. Observation 1
43 Observation Inference White precipitate/ppt persists on boiling SO42- ions Observation 2 Observation Inference White precipitate/ppt dissolves on boiling Cl - ions .(b)To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI). Observation 1 Observation Inference (i)acidified potassium manganate(VII)decolorized (ii)Orange colour of acidified potassium dichromate(VI) turns to green SO32- ions Observation 2 Observation Inference (i)acidified potassium manganate(VII) not decolorized (ii)Orange colour of acidified potassium dichromate(VI) does not turns to green CO32- ions Experiments/Observations: (b)Using Barium(II)nitrate(V)/ Barium(II)chloride I.
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Observation 1 Observation Inference White precipitate/ppt persists SO42- , Cl- ions Observation 2 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.(a)To the preserved sample observation 1 in (II) above, Heat to boil. Observation 1
43 Observation Inference White precipitate/ppt persists on boiling SO42- ions Observation 2 Observation Inference White precipitate/ppt dissolves on boiling Cl - ions .(b)To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI). Observation 1 Observation Inference (i)acidified potassium manganate(VII)decolorized (ii)Orange colour of acidified potassium dichromate(VI) turns to green SO32- ions Observation 2 Observation Inference (i)acidified potassium manganate(VII) not decolorized (ii)Orange colour of acidified potassium dichromate(VI) does not turns to green CO32- ions Experiments/Observations: (b)Using Barium(II)nitrate(V)/ Barium(II)chloride I. To about 5cm3 of a salt solution in a test tube add four drops of Barium(II) nitrate (V) / Barium(II)chloride. Preserve. Observation Inference White precipitate/ppt SO42- , SO32- , CO32- ions II. To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve. Observation 1 Observation Inference White precipitate/ppt persists SO42- , ions Observation 2
44 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI).
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To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve. Observation 1 Observation Inference White precipitate/ppt persists SO42- , ions Observation 2
44 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI). Observation 1 Observation Inference (i)acidified potassium manganate(VII)decolorized (ii)Orange colour of acidified potassium dichromate(VI) turns to green SO32- ions Observation 2 Observation Inference (i)acidified potassium manganate(VII) not decolorized (ii)Orange colour of acidified potassium dichromate(VI) does not turns to green CO32- ions Explanations Using Lead(II)nitrate(V) (i)Lead(II)nitrate(V) solution reacts with chlorides(Cl-), Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Lead(II)chloride, Lead(II)sulphate(VI), Lead(II) sulphate (IV) and Lead(II)carbonate(IV). Chemical/ionic equation: Pb2+(aq) + Cl- (aq) -> PbCl2(s) Pb2+(aq) + SO42+ (aq) -> PbSO4 (s) Pb2+(aq) + SO32+ (aq) -> PbSO3 (s) Pb2+(aq) + CO32+ (aq) -> PbCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and thus their white precipitates remain/ persists. 45 - Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. .
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Chemical/ionic equation: Pb2+(aq) + Cl- (aq) -> PbCl2(s) Pb2+(aq) + SO42+ (aq) -> PbSO4 (s) Pb2+(aq) + SO32+ (aq) -> PbSO3 (s) Pb2+(aq) + CO32+ (aq) -> PbCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and thus their white precipitates remain/ persists. 45 - Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. . Chemical/ionic equation: PbSO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + SO2 (g) PbCO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + CO2 (g) (iii)When Lead(II)chloride and Lead(II)sulphate(VI) are heated/warmed; - Lead(II)chloride dissolves in hot water/on boiling(recrystallizes on cooling) - Lead(II)sulphate(VI) do not dissolve in hot water thus its white precipitate persists/remains on heating/boiling. (iv)When sulphur(IV)oxide and carbon(IV)oxide gases are produced; - sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not.
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Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not. Chemical equation: Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l) These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water. Using Barium(II)nitrate(V)/ Barium(II)Chloride (i)Barium(II)nitrate(V) and/ or Barium(II)chloride solution reacts with Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Barium(II)sulphate(VI), Barium(II) sulphate (IV) and Barium(II)carbonate(IV). 46 Chemical/ionic equation: Ba2+(aq) + SO42+ (aq) -> BaSO4 (s) Ba2+(aq) + SO32+ (aq) -> BaSO3 (s) Ba2+(aq) + CO32+ (aq) -> BaCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists. - Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. .
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46 Chemical/ionic equation: Ba2+(aq) + SO42+ (aq) -> BaSO4 (s) Ba2+(aq) + SO32+ (aq) -> BaSO3 (s) Ba2+(aq) + CO32+ (aq) -> BaCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists. - Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. . Chemical/ionic equation: BaSO3 (s) + 2H+(aq) -> H2 O (l) + Ba2+(aq) + SO2 (g) BaCO3 (s) + 2H+(aq) -> H2 O (l) + Ba2+(aq) + CO2 (g) (iii) When sulphur(IV)oxide and carbon(IV)oxide gases are produced; - sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not. Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not.
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Carbon(IV)oxide will not. Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not. Chemical equation: Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l) These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water. 47 Summary test for Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts Practice revision question 1.
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Sulphur(IV)oxide will not. Chemical equation: Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l) These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water. 47 Summary test for Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts Practice revision question 1. Study the flow chart below and use it to answer the questions that follow Unknown salt Lead(II)nitrate(V) White precipitates of Cl-, SO42- , SO32- and CO32- Dilute nitric(V) acid white ppt dissolves in SO32- and CO32- white ppt persist /remains in SO32- and CO32- Acidified KMnO4 decolorized in SO32- White ppt with lime water in CO32- Acidified KMnO4 K2Cr2O7 / Lime water White ppt dissolves on heating in Cl- White ppt persist on heating in SO42- 2Heat to boil Sodium salt solution White precipitate Barium nitrate(VI) Colourless solution B Gas G and colour of solution changes orange to green Acidified K2Cr2O7 Dilute HCl
48 (a)Identify the: I: Sodium salt solution Sodium sulphate(IV)/Na2SO3 II: White precipitate Barium sulphate(IV)/BaSO3 III: Gas G Sulphur (IV)Oxide /SO2 IV: Colourless solution H Barium chloride /BaCl2 (b)Write an ionic equation for the formation of: I.White precipitate Ionic equation Ba2+(aq) + SO32-(aq) -> BaSO3(s) II.Gas G Ionic equation BaSO3(s)+ 2H+(aq) -> SO2 (g) + H2O (l) + Ba2+(aq) III. Green solution from the orange solution 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) 2. Study the flow chart below and answer the questions that follow.
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Study the flow chart below and use it to answer the questions that follow Unknown salt Lead(II)nitrate(V) White precipitates of Cl-, SO42- , SO32- and CO32- Dilute nitric(V) acid white ppt dissolves in SO32- and CO32- white ppt persist /remains in SO32- and CO32- Acidified KMnO4 decolorized in SO32- White ppt with lime water in CO32- Acidified KMnO4 K2Cr2O7 / Lime water White ppt dissolves on heating in Cl- White ppt persist on heating in SO42- 2Heat to boil Sodium salt solution White precipitate Barium nitrate(VI) Colourless solution B Gas G and colour of solution changes orange to green Acidified K2Cr2O7 Dilute HCl
48 (a)Identify the: I: Sodium salt solution Sodium sulphate(IV)/Na2SO3 II: White precipitate Barium sulphate(IV)/BaSO3 III: Gas G Sulphur (IV)Oxide /SO2 IV: Colourless solution H Barium chloride /BaCl2 (b)Write an ionic equation for the formation of: I.White precipitate Ionic equation Ba2+(aq) + SO32-(aq) -> BaSO3(s) II.Gas G Ionic equation BaSO3(s)+ 2H+(aq) -> SO2 (g) + H2O (l) + Ba2+(aq) III. Green solution from the orange solution 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) 2. Study the flow chart below and answer the questions that follow. 49 (i)Write equation for the reaction taking place at: I.The roasting furnace (1mk) 2FeS2 (s) + 5O2 (g) -> 2FeO(s) + 4SO2 (g) II.The absorption tower (1mk) H2SO4 (l) + SO3 (g) -> H2S2O7(l) III.The diluter (1mk) H2S2O7(l) + H2 O(l) -> 2H2SO4 (l) (ii)The reaction taking place in chamber K is SO2 (g) + 1/2O2 (g) SO3 (g) I.
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Green solution from the orange solution 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) 2. Study the flow chart below and answer the questions that follow. 49 (i)Write equation for the reaction taking place at: I.The roasting furnace (1mk) 2FeS2 (s) + 5O2 (g) -> 2FeO(s) + 4SO2 (g) II.The absorption tower (1mk) H2SO4 (l) + SO3 (g) -> H2S2O7(l) III.The diluter (1mk) H2S2O7(l) + H2 O(l) -> 2H2SO4 (l) (ii)The reaction taking place in chamber K is SO2 (g) + 1/2O2 (g) SO3 (g) I. Explain why it is necessary to use excess air in chamber K To ensure all the SO2 reacts II.Name another substance used in chamber K Vanadium(V)oxide 3.(a)Describe a chemical test that can be used to differentiate between sodium sulphate (IV) and sodium sulphate (VI). Add acidified Barium nitrate(V)/chloride.
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49 (i)Write equation for the reaction taking place at: I.The roasting furnace (1mk) 2FeS2 (s) + 5O2 (g) -> 2FeO(s) + 4SO2 (g) II.The absorption tower (1mk) H2SO4 (l) + SO3 (g) -> H2S2O7(l) III.The diluter (1mk) H2S2O7(l) + H2 O(l) -> 2H2SO4 (l) (ii)The reaction taking place in chamber K is SO2 (g) + 1/2O2 (g) SO3 (g) I. Explain why it is necessary to use excess air in chamber K To ensure all the SO2 reacts II.Name another substance used in chamber K Vanadium(V)oxide 3.(a)Describe a chemical test that can be used to differentiate between sodium sulphate (IV) and sodium sulphate (VI). Add acidified Barium nitrate(V)/chloride. White precipitate formed with sodium sulphate (VI) No white precipitate formed with sodium sulphate (IV) (b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas at s.t.p =22.4dm3) Chemical equation Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g) Mole ratio Cu(s: SO2 (g) = 1:1 Method 1 1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3 (120 x 1000) g -> (120 x 1000) g x 22.4.dm3) 63.5 g = 42330.7087
50 Method 2 Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles 63.5 Moles SO2 = Moles of Cu = 1889.7639 moles Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4) = 42330.7114 [email protected] solution TWhite precipitate RAcidified K2Cr2O7Barium nitrate(V)Colourless gasVDilute nitric (V) acid4.Use the reaction scheme below to answer the questions that follow.
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Explain why it is necessary to use excess air in chamber K To ensure all the SO2 reacts II.Name another substance used in chamber K Vanadium(V)oxide 3.(a)Describe a chemical test that can be used to differentiate between sodium sulphate (IV) and sodium sulphate (VI). Add acidified Barium nitrate(V)/chloride. White precipitate formed with sodium sulphate (VI) No white precipitate formed with sodium sulphate (IV) (b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas at s.t.p =22.4dm3) Chemical equation Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g) Mole ratio Cu(s: SO2 (g) = 1:1 Method 1 1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3 (120 x 1000) g -> (120 x 1000) g x 22.4.dm3) 63.5 g = 42330.7087
50 Method 2 Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles 63.5 Moles SO2 = Moles of Cu = 1889.7639 moles Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4) = 42330.7114 [email protected] solution TWhite precipitate RAcidified K2Cr2O7Barium nitrate(V)Colourless gasVDilute nitric (V) acid4.Use the reaction scheme below to answer the questions that follow. (a)Identify the: (i)cation responsible for the green solution T Cr3+ (ii)possible anions present in white precipitate R CO32-, SO32-, SO42- (b)Name gas V Sulphur (IV)oxide (c)Write a possible ionic equation for the formation of white precipitate R.
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Add acidified Barium nitrate(V)/chloride.
White precipitate formed with sodium sulphate (VI) No white precipitate formed with sodium sulphate (IV) (b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas at s.t.p =22.4dm3) Chemical equation Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g) Mole ratio Cu(s: SO2 (g) = 1:1 Method 1 1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3 (120 x 1000) g -> (120 x 1000) g x 22.4.dm3) 63.5 g = 42330.7087
50 Method 2 Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles 63.5 Moles SO2 = Moles of Cu = 1889.7639 moles Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4) = 42330.7114 [email protected] solution TWhite precipitate RAcidified K2Cr2O7Barium nitrate(V)Colourless gasVDilute nitric (V) acid4.Use the reaction scheme below to answer the questions that follow.
(a)Identify the: (i)cation responsible for the green solution T Cr3+ (ii)possible anions present in white precipitate R CO32-, SO32-, SO42- (b)Name gas V Sulphur (IV)oxide (c)Write a possible ionic equation for the formation of white precipitate R.
Ba2+ (aq) + CO32- (aq) -> BaCO3(s) Ba2+ (aq) + SO32- (aq) -> BaSO3(s) Ba2+ (aq) + SO42- (aq) -> BaSO4(s)
st
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