Chapter 9, Problem 21A

Interpretation Introduction

Interpretation:

The amount of ammonia gas formed when 1.25 g of ammonium carbonate decomposes needs to be deduced based on the given reaction.

Concept Introduction:

• A chemical reaction is represented in terms of a chemical equation with the reactants on the left and the products on the right.

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

• 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 21A

Mass of NH₃ = 0.442 g

Explanation of Solution

The given reaction is:

  ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{CO}}_{\text{3}}\text{(s) }\to {\text{NH}}_{\text{3}}{\text{(g) + CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(g)}$

The balanced equation is:

  ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{CO}}_{\text{3}}\text{(s) }\to 2{\text{NH}}_{\text{3}}{\text{(g) + CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(g)}$

Step 1: Calculate the moles of Fe present:

Mass of (NH₄)₂CO₃ present = 1.25 g

Molar mass of (NH₄)₂CO₃= 96.09 g/mole

  $\begin{array}{l}\text{Moles of }{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}\text{ = }\frac{\text{Mass of }{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{Molar Mass }{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}}\\ =\frac{\text{1}\text{.25 g}}{96.09\text{ g/mol}}=0.0130\text{ moles}\end{array}$

Step 2: Calculate the moles of NH₃ formed:

Based on the reaction stoichiometry:

1 mole of ammonium carbonate forms 2 moles of NH3

Therefore, 0.0130 moles of ammonium carbonate would yield:

  $\frac{0.0130{\text{ moles (NH}}_{\text{4}}{\text{)}}_{\text{2}}{\text{CO}}_{\text{3}}\times 2{\text{ moles NH}}_{\text{3}}}{1{\text{ mole (NH}}_{\text{4}}{\text{)}}_{\text{2}}{\text{CO}}_{\text{3}}}=0.026{\text{ moles NH}}_{\text{3}}$

Step 3: Calculate the mass of NH₃ formed:

Moles of NH₃ formed = 0.026

Molecular weight of NH₃ = 17 g/mol

  $\begin{array}{l}{\text{Mass of NH}}_{\text{3}}\text{ = Moles }\times \text{ Molecular weight}\\ =0.026\text{ moles }\times \text{ 17 g/mol = 0}\text{.442 g}\end{array}$

Conclusion

Therefore, mass of NH₃ formed is around 0.442 g.