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## Q. 5.12

Sodium azide ($NaN_{3}$) is used in some automobile air bags. The impact of a collision triggers the decomposition of $NaN_{3}$ as follows:

$2NaN_{3}(s) → 2Na(s) + 3N_{2}(g)$

The nitrogen gas produced quickly inflates the bag between the driver and the windshield and dashboard. Calculate the volume of N2 generated at 80°C and 823 mmHg by the decomposition of 60.0 g of $NaN_{3}$.

Strategy From the balanced equation, we see that $2 mol NaN_{3} \bumpeq 3 mol N_{2}$ so the conversion factor between $NaN_{3} and N_{2}$  is

$\frac{3 mol N_{2}}{2 mol NaN_{3}}$

Because the mass of $NaN_{3}$ is given, we can calculate the number of moles of $NaN_{3}$ and hence the number of moles of $N_{2}$ produced. Finally, we can calculate the volume of $N_{2}$ using the ideal gas equation.

## Verified Solution

First, we calculate the number of moles of $N_{2}$ produced by 60.0 g $NaN_{3}$ using the following sequence of conversions

grams of $NaN_{3}$ → moles of $NaN_{3}$ → moles of $N_{2}$

so that

moles of $N_{2}=60.0 \cancel{g NaN_{3}}\times \frac{1 \cancel{mol NaN_{3}}}{65.02 \cancel{g NaN_{3}}} \times \frac{3 mol N_{2}}{2 \cancel{mol NaN_{3}}}$

=1.38 mol  $N_{2}$

The volume of 1.38 moles of $N_{2}$ can be obtained by using the ideal gas equation

$V=\frac{nRT}{P}=\frac{(1.38 mol)(0.0821 L . atm/K . mol)(80+273 K)}{(823/760) atm}$

=36.9 L