Chapter 11.2, Problem 15PP

Interpretation Introduction

Interpretation:

The mass of ${\text{N}}_{\text{2}}$ produced from the decomposition of $100.0\text{ g}$ of ${\text{NaN}}_3$ needs to be calculated.

Concept introduction:

The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

Answer to Problem 15PP

The mass of ${\text{N}}_{\text{2}}$ produced from the decomposition of $100.0\text{ g}$ of ${\text{NaN}}_3$ is $28.85\text{ g}$.

Explanation of Solution

The given reaction used to inflate automobile bags is:

${\text{2NaN}}_3\text{(s) }\to {\text{ 2Na(s) + 3N}}_{\text{2}}\text{(g)}$

The number of moles of ${\text{NaN}}_3$ is calculated using formula:

$\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

The molar mass of ${\text{NaN}}_3$ is 65.01 g/mol.

Putting the values:

$\begin{array}{l}{\text{n}}_{{\text{NaN}}_3}\text{ = }\frac{\text{100}\text{.0 g}}{\text{65}\text{.01 g/mol}}\\ {\text{n}}_{{\text{NaN}}_3}\text{ = 1}\text{.54 mol}\end{array}$

From the balanced chemical reaction, the mole ratio of ${\text{NaN}}_3$ and ${\text{N}}_{\text{2}}$ is 2:3 that means 2 moles of ${\text{NaN}}_3$ produces 3 moles of ${\text{N}}_{\text{2}}$, so the number of moles of ${\text{N}}_{\text{2}}$ produced from 1.54 moles of ${\text{NaN}}_3$ will be:

$\text{1}{\text{.54 mol of NaN}}_3\text{ ×}\frac{{\text{2 mol of N}}_{\text{2}}}{{\text{3 mol of NaN}}_3}\text{ = 1}{\text{.03 mol of N}}_{\text{2}}$

Now, the mass of ${\text{N}}_{\text{2}}$ formed is calculated as:

The molar mass of ${\text{N}}_{\text{2}}$ is 28.01 g/mol. So,

$\begin{array}{l}{\text{Number of moles of N}}_{\text{2}}=\frac{{\text{Mass of N}}_{\text{2}}}{{\text{Molar mass of N}}_{\text{2}}}\\ \text{1}\text{.03 mol}=\frac{{\text{Mass of N}}_{\text{2}}}{\text{28}\text{.01 g/mol}}\\ {\text{Mass of N}}_{\text{2}}\text{ = }1.03\text{ mol}\times 28.01\text{ g/mol}\\ {\text{Mass of N}}_{\text{2}}\text{ = }28.85\text{ g}\end{array}$

Hence, the mass of ${\text{N}}_{\text{2}}$ produced from the decomposition of $100.0\text{ g}$ of ${\text{NaN}}_3$ is $28.85\text{ g}$.