Question

practice: gas stoichiometry
•automobile air bags inflate during a crash or sudden stop by the rapid generation of n₂(g) from sodium azide:
2 nan₃(s) → 2 na(s) + 3 n₂(g)
•how many grams of sodium azide are needed to produce sufficient n₂(g) to fill a 45 cm × 45 cm × 25 cm bag to a pressure of 1.20 atm at 15°c?

Explanation:

Step1: Calculate the volume of the bag

The volume \( V \) of the bag (rectangular prism) is \( V = l\times w\times h \).
Given \( l = 45\ cm\), \( w = 45\ cm\), \( h = 25\ cm\).
\( V=(45\times45\times25)\ cm^{3}=50625\ cm^{3}\).
Convert to liters: \( 1\ L = 1000\ cm^{3}\), so \( V=\frac{50625}{1000}\ L = 50.625\ L\).

Step2: Use the ideal gas law \( PV = nRT \) to find the moles of \( N_{2}\)

The ideal - gas law is \( PV=nRT \), where \( P = 1.20\ atm\), \( V = 50.625\ L\), \( R=0.0821\ L\cdot atm/(mol\cdot K)\), and \( T=(15 + 273)\ K=288\ K\).
Solve for \( n\): \( n=\frac{PV}{RT}\).
Substitute the values: \( n=\frac{1.20\ atm\times50.625\ L}{0.0821\ L\cdot atm/(mol\cdot K)\times288\ K}\).
\( n=\frac{60.75}{23.6448}\ mol\approx2.57\ mol\).

Step3: Use the stoichiometry of the reaction to find moles of \( NaN_{3}\)

From the balanced equation \( 2NaN_{3}(s)\to2Na(s)+3N_{2}(g)\), the mole ratio of \( NaN_{3}\) to \( N_{2}\) is \( \frac{2}{3}\).
If \( n_{N_{2}} = 2.57\ mol\), then \( n_{NaN_{3}}=\frac{2}{3}\times n_{N_{2}}\).
\( n_{NaN_{3}}=\frac{2}{3}\times2.57\ mol\approx1.71\ mol\).

Step4: Calculate the mass of \( NaN_{3}\)

The molar mass of \( NaN_{3}\) is \( M=(22.99+3\times14.01)\ g/mol=(22.99 + 42.03)\ g/mol = 65.02\ g/mol\).
Use the formula \( m=n\times M\).
\( m_{NaN_{3}}=n_{NaN_{3}}\times M\).
Substitute \( n_{NaN_{3}} = 1.71\ mol\) and \( M = 65.02\ g/mol\).
\( m_{NaN_{3}}=1.71\ mol\times65.02\ g/mol\approx111\ g\).

Answer:

\( 111\ g\)