Question

Sodium azide $\left(Na{N}_3\right)$ is an explosive chemical used in automobile airbags. It is made by the following reaction:
$NaN{O}_3+3NaN{H}_2\to Na{N}_3+3NaOH+N{H}_3$

One takes 17.0 g of $NaN{O}_3$ and 13.0 g of $NaN{H}_2$. If this amount is used in automobile bags, how much $N_2$ is formed due to collision at STP?

• A. 3.73 L
• B. 12.93 L
• C. 8.62 L
• D. 2.15 L

Answer 1

The correct option is A 3.73 L

$\underset{\underset{17g}{1\text{mol}}}{NaN{O}_3}+\underset{\underset{13g}{3\text{mol}}}{3NaN{H}_2}\to Na{N}_3+3NaOH+N{H}_3$

Finding the limiting reagent,

moles of $NaN{O}_3=\frac{17}{85}=0.2\text{mol}$ moles of $NaN{H}_2=\frac{13}{39}=\frac{1}{3}\text{mol}$

$\frac{\text{moles of}NaN{O}_3}{1}=\frac{0.2}{1}$
$\frac{\text{moles of}NaN{H}_2}{3}=\frac{1}{3\times 3}$

Hence, $NaN{H}_2$ is the limiting reactant.

Amount of $Na{N}_3$ is obtained from $NaN{H}_2$,
3 moles $NaN{H}_2$ give = 1 mole $Na{N}_3$
$\frac{1}{3}$ mole gives $=\frac{1}{3\times 3}\text{mol}Na{N}_3=\frac{1}{9}\text{mole}Na{N}_3$

$2Na{N}_3\to 3N_2$

2 moles $Na{N}_3$ gives = 3 mole $N_2$ gas

$\frac{1}{9}$ mole $Na{N}_3$ give $=\frac{3}{2\times 9}\text{mole}{N}_2=\frac{1}{6}\text{mole}{N}_2$
1 mole $N_2$ at STP = 22.4 L
$\therefore \frac{1}{6}\text{mole}{N}_2$ at STP = 3.73 L