Question

8. What mass of NaN3 is needed to inflate an 18.0 L air bag in an automobile at 25.0°C and 1.00 atm according to the following reaction? 6NaN3(s) + Fe2O3(s) + 3Na2O(s) + 2Fe(s) + 9N2(9)

Answer 1

The number of moles (n) of nitrogen gas are related to pressure (P), volume (V) and temperature (T) by ideal gas equation

$n=\frac{P \times V}{R \times T}$

Here, R is ideal gas constant.

Substitute values in above equation

$n=\frac{1.00 \ atm \times 18.0 \ L}{0.08206 \ \frac{L atm}{mol K} \times \left ( 25.0+273.15 \right ) \ K}$

$n=0.736 \ mol$

Using reaction stoichiometry, convert number of moles of nitrogen into number of moles of sodiumazide

$0.736 \ mol \ N_2 \times \frac{6 \ mol \ NaN_3}{9 \ mol \ N_2}=0.490 \ mol \ NaN_3$

The molar mass of $NaN_3=23+3(14)=65 \ g/mol$

Mass of $NaN_3= 65 \ g/mol \times 0.490 \ mol =31.9 \ g$