Determine the theoretical yield of H2S (in moles) if 4.0 mol Al2S3 and 4.0 mol H2O are reacted according to the following balanced reaction. A possibly useful molar mass is Al2S3 = 150.17 g/mol.
Determine the theoretical yield of H2S (in moles) if 4.0 mol Al2S3 and 4.0 mol H2O are reacted according to the following balanced reaction. A possibly useful molar mass is Al2S3 = 150.17.I am having a little trouble with this one.nswers. Best Answer: You need 6 moles of H2O for every mole of Al2S3 but you have equal moles so the limiting factor with be the amount of H2O. From the equation 6 moles H2O gives 3 moles H2S So 4 moles H2O will give 2 moles H2S. Assignment Help. Determine the theoretical yield of H2S (in moles) if 4.0 mol Al2S3 and 4.0 mol H2O are reacted according to the following balanced reaction.A possibly useful molar mass is Al2S3 = 150.17 g/mol.ow many moles of CO(NH2)2 are present in 150 g of water, if the freezing point of the solution is -4.02C? kf(water) = 1.86 K · kg/mol. Determine the theoretical yield of H2S (in moles) if 4.0 mol Al2S3 and 4.0 mol H2O are reacted according to the following balanced reaction. A possibly useful molar mass is Al2S3 = 150.17 g/mol. Al2S3(s) + 6 H2O(l) ? 2 Al(OH)3(s) + 3 H2S(g) 18 mol H2S 6.0 mol H2S 2.0 mol H2S 4.0 mol H2S 12 mol H2S.etermine the theoretical yield of H2S (in moles) if 4.0 mol Al2S3 and 4.0 mol H2O are reacted according to the following balanced reaction. A possibly useful molar mass is Al2S3 = 150.17 g/mol. Al2S3(s) + 6 H2O(l) ? 2 Al(OH)3(s) + 3 H2S(g) 18 mol H2S 6.0 mol H2S 2.0 mol H2S 4.0 mol H2S 12 mol H2S. 13) Determine the theoretical yield of HCl if 60.0 g of BCl3 and 37.5 g of H2O are reacted according to the following balanced reaction. A possibly useful molar mass is BCl3 = 117.16 g/mol.BCl3(g) + 3 H2O(l) → H3BO3(s) + 3 HCl(g). A) 75.9 g HCl B) 132 g HCl C) 187 g HCl D) 56.0 g HCl E) 25.3 g HCl Answer: D. 14) Determine the theoretical yield of H2S (in moles) if 4.0 mol Al2S3 and 4.0 mol H2O are reacted according to the following balanced reaction.A possibly useful molar mass is Al2S3 = 150.17 g/mol.he molar masses are as follows: H2O = 18.02 g/mol, HNO3 = 63.02 g/mol. 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g). A) 38.0 g H2O B) 21.7 g H2O C) 43.4 g H2O D) 10.9 g H2O E) 26.5 g H2O Answer: D. 5) Consider the following balanced reaction. Determine the theoretical yield of HCl if 60.0 g of BCl3 and 37.5 g of H2O are reacted according to the following balanced reaction.A possibly useful molar mass is BCl3 = 117.16 g/mol.etermine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N2O4 and 45.0 g N2H4. Some. possibly useful molar masses are as follows: N2O4 = 92.02 g/mol, N2H4 = 32.05 g/mol. Answer: C. 17) Determine the theoretical yield of HCl if 60.0 g of BCl3 and 37.5 g of H2O are reacted according to the following balanced reaction. A possibly useful molar mass is BCl3 = 117.16 g/mol.he first sample produced 13.0 g of potassium and 42.3 g of iodine. If the second sample produced 24.4 kg of potassium, how many kg of iodine were produced? A) 13.3 kg B) 22.5 kg C) 79.4 kg D) 44.4 kg E) 92.4 kg Answer: C A 14.01 g sample of N2 reacts with 3.02 g of H2 to form ammonia (NH3). I'd appreciate it if someone could show me how to do this problem step by step. Thanks. Determine the theoretical yield of HCl if 60.0 g of BCl3 and 37.5 g of H2O are reacted according to the following balanced reaction. A possibly useful molar mass is BCl3 = 117.16 g/mol.Note: if there were less than 1.5363 moles of H2O, H2O would be the limiting reactant) All of the BCl3 reacts, so 3*0.5121=1.5363 moles of HCl are formed Mass of HCl: 1.5363mol*36.45g/mol= 56 g HCl. 17) Determine the theoretical yield of HCl if 60.0 g of BCl3 and 37.5 g of H2O are reacted according to the following balanced reaction.he first sample produced 13.0 g of potassium and 42.3 g of iodine. If the second sample produced 24.4 kg of potassium, how many kg of iodine were produced? A) 13.3 kg B) 22.5 kg C) 79.4 kg D) 44.4 kg E) 92.4 kg Answer: C A 14.01 g sample of N2 reacts with 3.02 g of H2 to form ammonia (NH3). Answers. Best Answer: You need 6 moles of H2O for every mole of Al2S3 but you have equal moles so the limiting factor with be the amount of H2O. From the equation 6 moles H2O gives 3 moles H2S So 4 moles H2O will give 2 moles H2S.Source(s):nswers. Best Answer: You need 6 moles of H2O for every mole of Al2S3 but you have equal moles so the limiting factor with be the amount of H2O. From the equation 6 moles H2O gives 3 moles H2S So 4 moles H2O will give 2 moles H2S. The molar mass of KNO3 is 101.11 g/mol. 4 KNO3(s) → 2 K2O(s) + 2 N2(g) + 5 O2(g) A) 0.290 mol N2 B) 0.580 mol N2 C) 18.5 mol N2 D) 0.724 mol N2 E) 1.73 mol N2 Answer: A Two samples of calcium fluoride are decomposed into their constituent elements.The first sample produced 0.154 g of calcium and 0.146 g of fluorine.he first sample produced 13.0 g of potassium and 42.3 g of iodine. If the second sample produced 24.4 kg of potassium, how many kg of iodine were produced? A) 13.3 kg B) 22.5 kg C) 79.4 kg D) 44.4 kg E) 92.4 kg Answer: C A 14.01 g sample of N2 reacts with 3.02 g of H2 to form ammonia (NH3).
You need 6 moles of H2O for every mole of Al2S3 but you have equal moles so the limiting factor with be the amount of H2O. From the equation 6 moles H2O gives 3 moles H2S So 4 moles H2O will give 2 moles H2S.