Question

part a
if an automobile air bag has a volume of 11.9 l, what mass of nan₃ (in g) is required to fully inflate the air bag upon impact? assume stp conditions.
express your answer in grams to three significant figures.
view available hint(s)

m =

g nan₃

Explanation:

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

At STP (\(T = 273\space K\), \(P= 1\space atm\)), the molar volume of an ideal gas is \(V_m=22.4\space L/mol\).
We use the formula \(n=\frac{V}{V_m}\), where \(V = 11.9\space L\) and \(V_m = 22.4\space L/mol\).
So, \(n=\frac{11.9}{22.4}\space mol\).

Step2: Write the balanced chemical equation

The decomposition reaction of \(NaN_3\) is \(2NaN_3(s)
ightarrow 2Na(s)+3N_2(g)\).
From the equation, the mole ratio of \(NaN_3\) to \(N_2\) is \(n(NaN_3):n(N_2)=2:3\).
So, \(n(NaN_3)=\frac{2}{3}n(N_2)\).
Substituting \(n(N_2)=\frac{11.9}{22.4}\space mol\), we get \(n(NaN_3)=\frac{2}{3}\times\frac{11.9}{22.4}\space mol\).

Step3: Calculate the molar mass of \(NaN_3\)

The molar mass of \(NaN_3\), \(M(NaN_3)=(22.99 + 3\times14.01)\space g/mol=65.02\space g/mol\).

Step4: Calculate the mass of \(NaN_3\)

We use the formula \(m = n\times M\).
\(m(NaN_3)=n(NaN_3)\times M(NaN_3)\)
\(m(NaN_3)=\frac{2}{3}\times\frac{11.9}{22.4}\times65.02\space g\)

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Answer:

\(23.0\space g\)