Question

automobile air bags inflate following a serious impact. the impact triggers the chemical reaction:
2nan₃(s)→2na(s)+3n₂(g)
if an automobile air bag has a volume of 11.8 l, what mass of nan₃ (in g) is required to fully inflate the air bag upon impact? assume stp conditions.

Explanation:

Step1: Calculate the number of moles of \(N_2\) gas

At STP (\(T = 273\space K\), \(P= 1\space atm\)), the molar volume of an ideal gas \(V_m=22.4\space L/mol\).
We use the formula \(n=\frac{V}{V_m}\), where \(V = 11.8\space L\) and \(V_m = 22.4\space L/mol\)

$$n_{N_2}=\frac{11.8\space L}{22.4\space L/mol}\approx0.527\space mol$$

Step2: Use the stoichiometry of the chemical reaction

From the balanced chemical equation \(2NaN_3(s)\to2Na(s)+3N_2(g)\), the mole ratio of \(NaN_3\) to \(N_2\) is \(\frac{n_{NaN_3}}{n_{N_2}}=\frac{2}{3}\)

$$n_{NaN_3}=\frac{2}{3}\times n_{N_2}$$

Substitute \(n_{N_2}=0.527\space mol\)

$$n_{NaN_3}=\frac{2}{3}\times0.527\space mol\approx0.351\space mol$$

Step3: Calculate the mass of \(NaN_3\)

The molar mass of \(NaN_3\) (\(M\)): \(M = 23\space g/mol+(14\times3)\space g/mol=65\space g/mol\)
We use the formula \(m = n\times M\), where \(n = n_{NaN_3}=0.351\space mol\) and \(M = 65\space g/mol\)

$$m_{NaN_3}=0.351\space mol\times65\space g/mol = 22.8\space g$$

Answer:

\(22.8\space g\)