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  # Using the following information to help you answer parts (a) to (c):  $HN{O_3} + NaOH \to NaN{O_3} + {H_2}O$ RAM: $H = 1;Na = 23;O = 16;N = 14$ (A) If you have $0.55$ moles of $HN{O_3}$ how many oxygen atoms would you have?(B) If $0.365g$ of $NaOH$ is reacted with excess of $HN{O_3}$ what is the maximum mass of $NaN{O_3}$ which could be recovered?(C) $12.05c{m^3}$ of a $0.2065M$ aqueous $NaOH$ solution is titrated to the end-point by $25.05c{m^3}$ of $HN{O_3}$ solution. What is the concentration of the $HN{O_3}$ ?
  
  Last updated date: 23rd Apr 2024
  Total views: 283.8k
  Views today: 6.83k
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  283.8k+ views
  Hint: Moles is the concept that represents the number of atoms present in one mole, which is equal to $6.023 \times {10^{23}}$ known as Avogadro’s number. As one mole consists of Avogadro's number of atoms, by multiplying the given number of moles with Avogadro’s number gives the number of atoms.
  
  Given reaction is $HN{O_3} + NaOH \to NaN{O_3} + {H_2}O$
  One mole consists of $6.023 \times {10^{23}}$ number of molecules, $0.55$ moles of $HN{O_3}$ consists of $6.023 \times {10^{23}} \times 0.55 = 3.312 \times {10^{23}}$ molecules. As there are three oxygen atoms in one molecule of $HN{O_3}$ .
  The oxygen atoms will be $3 \times 3.312 \times {10^{23}} = 9.93 \times {10^{23}}$
  Thus, $0.55$ moles of $HN{O_3}$ consist of $9.93 \times {10^{23}}$ oxygen atoms.
  The number of moles is the ratio of mass to molar mass.
  Given that the $0.365g$ of $NaOH$ is reacted, the molar mass of $NaOH$ is $40gmo{l^{ - 1}}$ .
  The number of moles of $NaOH$ will be $\dfrac{{0.365}}{{40}} = 0.091moles$
  As one mole of $NaOH$ reacts with one mole of $HN{O_3}$ to produce one mole of $NaN{O_3}$
  Thus, $0.091moles$ of $NaOH$ produces $0.091moles$ of $NaN{O_3}$
  The molar mass of $NaN{O_3}$ is $85gmo{l^{ - 1}}$
  Thus, the mass of $NaN{O_3}$ will be $0.091mol \times 85gmo{l^{ - 1}} = 0.7335g$
  If $0.365g$ of $NaOH$ is reacted with excess of $HN{O_3}$ the maximum mass of $NaN{O_3}$ which could be recovered is $0.7335g$
  Given that the initial volume is $12.05c{m^3}$ with initial concentration of $0.2065M$ aqueous $NaOH$ solution is titrated to the end-point by a final volume of $25.05c{m^3}$ of $HN{O_3}$ solution.
  Substitute these values in the formula, ${M_1}{V_1} = {M_2}{V_2}$
  $12.05c{m^3} \times 0.2065M = {M_2} \times 25.05c{m^3}$
  Further simplification, ${M_2} = 0.099M$
  Thus, $12.05c{m^3}$ of a $0.2065M$ aqueous $NaOH$ solution is titrated to the end-point by $25.05c{m^3}$ of $HN{O_3}$ solution the concentration of the $HN{O_3}$ will be $0.099M$ .
  
  Note:
  While calculating the number of moles, the molar mass should be taken exactly and the balanced chemical equation only must be considered. Based on this balanced equation only, the mole ratio is taken and according to this the moles of product can be calculated.
explanation: The document contains a series of chemistry questions and their corresponding answers. The questions are related to stoichiometry and chemical reactions, specifically involving the reaction between nitric acid (HN₃) and sodium hydroxide (NaOH). The questions ask for the number of oxygen atoms in a given amount of HN₃, the maximum mass of NaN₃ that can be produced from a given amount of NaOH, and the concentration of HN₃ in a titration experiment. Each question is followed by a detailed step-by-step solution.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
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  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
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  # Using the following information to help you answer parts (a) to (c):  $HN{O_3} + NaOH \to NaN{O_3} + {H_2}O$ RAM: $H = 1;Na = 23;O = 16;N = 14$ (A) If you have $0.55$ moles of $HN{O_3}$ how many oxygen atoms would you have?(B) If $0.365g$ of $NaOH$ is reacted with excess of $HN{O_3}$ what is the maximum mass of $NaN{O_3}$ which could be recovered?(C) $12.05c{m^3}$ of a $0.2065M$ aqueous $NaOH$ solution is titrated to the end-point by $25.05c{m^3}$ of $HN{O_3}$ solution. What is the concentration of the $HN{O_3}$ ?
  
  Last updated date: 23rd Apr 2024
  Total views: 283.8k
  Views today: 6.83k
  Verified
  283.8k+ views
  Hint: Moles is the concept that represents the number of atoms present in one mole, which is equal to $6.023 \times {10^{23}}$ known as Avogadro’s number. As one mole consists of Avogadro's number of atoms, by multiplying the given number of moles with Avogadro’s number gives the number of atoms.
  
  Given reaction is $HN{O_3} + NaOH \to NaN{O_3} + {H_2}O$
  One mole consists of $6.023 \times {10^{23}}$ number of molecules, $0.55$ moles of $HN{O_3}$ consists of $6.023 \times {10^{23}} \times 0.55 = 3.312 \times {10^{23}}$ molecules. As there are three oxygen atoms in one molecule of $HN{O_3}$ .
  The oxygen atoms will be $3 \times 3.312 \times {10^{23}} = 9.93 \times {10^{23}}$
  Thus, $0.55$ moles of $HN{O_3}$ consist of $9.93 \times {10^{23}}$ oxygen atoms.
  The number of moles is the ratio of mass to molar mass.
  Given that the $0.365g$ of $NaOH$ is reacted, the molar mass of $NaOH$ is $40gmo{l^{ - 1}}$ .
  The number of moles of $NaOH$ will be $\dfrac{{0.365}}{{40}} = 0.091moles$
  As one mole of $NaOH$ reacts with one mole of $HN{O_3}$ to produce one mole of $NaN{O_3}$
  Thus, $0.091moles$ of $NaOH$ produces $0.091moles$ of $NaN{O_3}$
  The molar mass of $NaN{O_3}$ is $85gmo{l^{ - 1}}$
  Thus, the mass of $NaN{O_3}$ will be $0.091mol \times 85gmo{l^{ - 1}} = 0.7335g$
  If $0.365g$ of $NaOH$ is reacted with excess of $HN{O_3}$ the maximum mass of $NaN{O_3}$ which could be recovered is $0.7335g$
  Given that the initial volume is $12.05c{m^3}$ with initial concentration of $0.2065M$ aqueous $NaOH$ solution is titrated to the end-point by a final volume of $25.05c{m^3}$ of $HN{O_3}$ solution.
  Substitute these values in the formula, ${M_1}{V_1} = {M_2}{V_2}$
  $12.05c{m^3} \times 0.2065M = {M_2} \times 25.05c{m^3}$
  Further simplification, ${M_2} = 0.099M$
  Thus, $12.05c{m^3}$ of a $0.2065M$ aqueous $NaOH$ solution is titrated to the end-point by $25.05c{m^3}$ of $HN{O_3}$ solution the concentration of the $HN{O_3}$ will be $0.099M$ .
  
  Note:
  While calculating the number of moles, the molar mass should be taken exactly and the balanced chemical equation only must be considered. Based on this balanced equation only, the mole ratio is taken and according to this the moles of product can be calculated.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a series of chemistry questions and their corresponding answers. The questions are related to stoichiometry and chemical reactions, specifically involving the reaction between nitric acid (HN₃) and sodium hydroxide (NaOH). The questions ask for the number of oxygen atoms in a given amount of HN₃, the maximum mass of NaN₃ that can be produced from a given amount of NaOH, and the concentration of HN₃ in a titration experiment. Each question is followed by a detailed step-by-step solution.
  
  Label: multiple-question-and-answer