Chapter 9, Problem 23A

Interpretation Introduction

Interpretation:

The amount of nitrogen, oxygen and water vapor formed when 1.25 g of ammonium nitrate decomposes needs to be deduced based on the given reaction.

Concept Introduction:

• A chemical reaction is expressed as a chemical equation having reactants and products on left and right side of the reaction arrow respectively.

  $\text{Reactants }\to \text{ Products}$

• The coefficient of a balanced chemical equation i.e. the stoichiometry gives the amount of reactants and products involved in the reaction.
• Chemical equations can therefore be used to determine the amount of products formed from a known quantity of reactants.

Answer to Problem 23A

Mass of N₂ = 0.437 g, Mass of O₂ = 0.499 g , Mass of water vapour =0.562 g.

Explanation of Solution

The given reaction is:

  ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{(s) }\to {\text{N}}_2{\text{(g) + O}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(g)}$

The balanced equation is:

  ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{(s) }\to {\text{N}}_2{\text{(g) + O}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{O(g)}$

Step 1: Calculate the moles of NH₄NO₃ present:

Mass of NH₄NO₃ present= 1.25 g

Molar mass of NH₄NO₃ = 80 g/mole

  $\begin{array}{l}{\text{Moles of NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{= }\frac{{\text{Mass of NH}}_{\text{4}}{\text{NO}}_{\text{3}}}{{\text{Molar Mass NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\\ =\frac{\text{1}\text{.25 g}}{80\text{ g/mol}}=0.0156\text{ moles}\end{array}$

Step 2: Calculate the mass of N₂ formed:

Based on the reaction stoichiometry:

1 mole of ammonium nitrate forms 1 mole of N₂ gas

Therefore, 0.0156 moles of ammonium nitrate would yield 0.0156 moles of N₂.

Molecular weight of N₂ = 28 g/mol

  $\begin{array}{l}{\text{Mass of N}}_2\text{ = Moles }\times \text{ Molecular weight}\\ =0.0156\text{ moles }\times \text{ 28 g/mol = 0}\text{.437 g}\end{array}$

Step 3: Calculate the mass of O₂ formed:

Based on the reaction stoichiometry:

1 mole of ammonium nitrate forms 1 mole of O₂ gas

Therefore, 0.0156 moles of ammonium nitrate would yield 0.0156 moles of O₂

Molecular weight of O₂ = 32 g/mol

  $\begin{array}{l}{\text{Mass of O}}_2\text{ = Moles }\times \text{ Molecular weight}\\ =0.0156\text{ moles }\times \text{ 32 g/mol = 0}\text{.499 g}\end{array}$

Step 4: Calculate the mass of water vapor formed:

Based on the reaction stoichiometry:

1 mole of ammonium nitrate forms 2 moles of H₂O gas

Therefore, 0.0156 moles of ammonium nitrate would yield:

  $\frac{0.0156{\text{ moles NH}}_{\text{4}}{\text{NO}}_{\text{3}}\times 2{\text{ moles H}}_{\text{2}}\text{O}}{1{\text{ mole NH}}_{\text{4}}{\text{NO}}_{\text{3}}}=0.0312{\text{ moles H}}_{\text{2}}\text{O}$

Molecular weight of H₂O = 18 g/mol

  $\begin{array}{l}{\text{Mass of H}}_2\text{O = Moles }\times \text{ Molecular weight}\\ =0.0312\text{ moles }\times \text{ 18 g/mol = 0}\text{.562 g}\end{array}$

Conclusion

Therefore, the amount of nitrogen, oxygen and water vapor formed are 0.437 g, 0.499 g and 0.562 g respectively.