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2. Automotive air bags inflate when sodium azide decomposes as shown in following equation. Calcuiate the volume of gas that forms at \( S T P \) when 35.0 g sodium azide decomposes \( P \vee=\cap R T \)

Ask by Lawson Johnson. in the United States
Mar 13,2025

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When 35.0 grams of sodium azide decompose, about 18.1 liters of nitrogen gas are produced at STP.

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When sodium azide (\( NaN_3 \)) decomposes, it produces nitrogen gas (\( N_2 \)), sodium (which is solid and not relevant for gas volume), and additional products as per the balanced reaction. The balanced equation is: \[ 2 \, NaN_3(s) \rightarrow 3 \, N_2(g) + 2 \, Na(s) \] To find the volume of gas produced at STP from the decomposition of 35.0 g of sodium azide, start by calculating the number of moles of \( NaN_3 \): 1. **Molar Mass Calculation**: - The molar mass of sodium azide (\( NaN_3 \)) is approximately \( 65.01 \, g/mol \). 2. **Moles of Sodium Azide**: \[ \text{Moles of } NaN_3 = \frac{35.0 \, g}{65.01 \, g/mol} \approx 0.538 \, mol \] 3. **Using the Balanced Equation**: From the balanced equation, 2 moles of \( NaN_3 \) produce 3 moles of \( N_2 \). Therefore: \[ 0.538 \, mol \, NaN_3 \times \frac{3 \, mol \, N_2}{2 \, mol \, NaN_3} \approx 0.807 \, mol \, N_2 \] 4. **Volume of Gas at STP**: At STP (Standard Temperature and Pressure, \( 0^\circ C \) and \( 1 \, atm \)), 1 mole of any ideal gas occupies 22.4 L. \[ \text{Volume} = 0.807 \, mol \times 22.4 \, L/mol \approx 18.1 \, L \] Thus, the volume of nitrogen gas produced when 35.0 g of sodium azide decomposes is approximately 18.1 liters at STP.

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