The decomposition of sodium azide (NaN3) solid to sodium liquid and nitrogen gas is used in air bags. What volume of nitrogen gas measured at 725 mm of Hg and 26°C is produced when 75.0g of sodium azide (molar mass = 65.01g/mol) decomposes.
The correct answer and explanation is :
Answer:
Volume of nitrogen gas produced = 33.9 liters
Explanation (Approx. 300 words):
The decomposition of sodium azide (NaN₃) is a key chemical reaction used in airbag systems. When triggered, it rapidly decomposes into sodium metal (Na) and nitrogen gas (N₂) according to the following balanced chemical equation:
$$
2 \text{NaN}_3(s) \rightarrow 2 \text{Na}(l) + 3 \text{N}_2(g)
$$
This tells us that 2 moles of NaN₃ yield 3 moles of N₂ gas.
Step 1: Calculate Moles of NaN₃
Given:
- Mass of NaN₃ = 75.0 g
- Molar mass of NaN₃ = 65.01 g/mol
$$
\text{Moles of NaN}_3 = \frac{75.0\text{ g}}{65.01\text{ g/mol}} \approx 1.1537 \text{ mol}
$$
Step 2: Use Mole Ratio to Find Moles of N₂
From the balanced equation:
$$
2 \text{ mol NaN}_3 \rightarrow 3 \text{ mol N}_2
$$
So,
$$
\text{Moles of N}_2 = 1.1537 \times \frac{3}{2} \approx 1.7306 \text{ mol}
$$
Step 3: Use Ideal Gas Law to Find Volume
The ideal gas law:
$$
PV = nRT
$$
Convert units:
- $P = 725 \text{ mmHg} = \frac{725}{760} \text{ atm} \approx 0.9539 \text{ atm}$
- $T = 26^\circ C = 299 \text{ K}$
- $R = 0.0821 \, \text{L·atm/mol·K}$
- $n = 1.7306 \text{ mol}$
Plug into the equation:
$$
V = \frac{nRT}{P} = \frac{1.7306 \times 0.0821 \times 299}{0.9539} \approx 33.9 \text{ L}
$$
Conclusion:
33.9 liters of nitrogen gas are produced when 75.0 g of NaN₃ decomposes under the specified conditions. This reaction’s rapid gas expansion is what inflates airbags in fractions of a second during vehicle collisions.