Question

Automobile airbags are inflated with \[{{\text{N}}_{\text{2}}}\] gas which is formed by the decomposition of solid sodium azide \[\left( {{\text{Na}}{{\text{N}}_{\text{3}}}} \right)\] . The other product is \[{\text{Na}}\] - metal. Calculate the volume of \[{{\text{N}}_{\text{2}}}\] gas at \[{27^o}{\text{C}}\]and \[{\text{1 atm}}\] formed by the decomposing of \[{\text{130 gm}}\] of sodium azide.

Answer 1

Hint:In this problem, you will need the ideal gas equation \[V = \dfrac{{n \times R \times T}}{P}\] to calculate the volume of nitrogen gas. For this, you can calculate the number of moles of nitrogen gas by using the reaction stoichiometry.

Complete answer:
The decomposition of sodium azide gives sodium metal and nitrogen gas. Write a balanced chemical equation for the above reaction as shown below:
\[{\text{Na}}{{\text{N}}_3}{\text{ }} \to {\text{ Na + }}\dfrac{3}{2}{\text{ }}{{\text{N}}_2}\]
The mass of sodium azide is one hundred and thirty grams and the molecular weight of sodium azide is sixty five grams per mole. Divide the mass of sodium azide with its molecular weight to obtain the number of moles of sodium azide.
\[\Rightarrow\text{Number of moles of sodium azide} \dfrac{{130}}{{65}} = 2\]
Hence, one hundred and thirty grams of sodium azide corresponds to two moles.
According to the above balanced chemical equation, when you decompose one mole of sodium azide, you will get \[\dfrac{3}{2}\] moles of nitrogen gas.
Thus, when you decompose two moles of sodium azide, you will get \[\dfrac{3}{2} \times 2 = 3\] moles of nitrogen gas.

Write the ideal gas equation
\[PV = nRT\]
Here, O is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant and T is absolute temperature.
Rearrange the ideal gas equation to obtain an expression in terms of V
\[V = \dfrac{{n \times R \times T}}{P}\]

Substitute 3 for n, 0.082 for R, 300 for T and 1 for P in the above expression
\[
  V = \dfrac{{n \times R \times T}}{P} \\
\Rightarrow V = \dfrac{{3{\text{ mol}} \times 0.082{\text{ L}}{\text{.atm/mol}}{\text{.K }} \times {\text{ 300 K}}}}{{1{\text{ atm}}}} \\
\Rightarrow V = 73.89{\text{ L}}
 \]
Hence, the volume of \[{{\text{N}}_{\text{2}}}\] gas at \[{27^o}{\text{C}}\]and \[{\text{1 atm}}\] formed by the decomposing of \[{\text{130 gm}}\] of sodium azide is \[73.89{\text{ L}}\] .

Note:
At STP, one mole of any gas occupies a volume of \[22.4{\text{ L}}\] . Here, STP refers to standard temperature of \[273{\text{ K}}\] and standard pressure of \[1{\text{ atm}}\] .