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question 21 4 pts automobile air bags use the decomposition of sodium a…

Question

question 21
4 pts
automobile air bags use the decomposition of sodium azide as their source of gas for rapid inflation:
2nan3(s)→2na(s)+3n2(g)
what mass of nan3 (in grams) is required to provide 40.0l of n2 at 25.0°c and 763 torr?

Explanation:

Step1: Convert pressure to atm and temperature to Kelvin

Pressure: \( 763 \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}} \approx 1.004 \text{ atm} \)
Temperature: \( T = 25.0^\circ \text{C} + 273.15 = 298.15 \text{ K} \)

Step2: Use ideal gas law to find moles of \( \text{N}_2 \)

Ideal gas law: \( PV = nRT \), so \( n = \frac{PV}{RT} \)
\( R = 0.0821 \text{ L·atm/(mol·K)} \)
\( n_{\text{N}_2} = \frac{1.004 \text{ atm} \times 40.0 \text{ L}}{0.0821 \text{ L·atm/(mol·K)} \times 298.15 \text{ K}} \approx \frac{40.16}{24.48} \approx 1.64 \text{ mol} \)

Step3: Use stoichiometry to find moles of \( \text{NaN}_3 \)

From the reaction: \( 2 \text{ mol } \text{NaN}_3
ightarrow 3 \text{ mol } \text{N}_2 \)
\( n_{\text{NaN}_3} = \frac{2}{3} \times n_{\text{N}_2} = \frac{2}{3} \times 1.64 \text{ mol} \approx 1.093 \text{ mol} \)

Step4: Calculate mass of \( \text{NaN}_3 \)

Molar mass of \( \text{NaN}_3 \): \( 22.99 + 3 \times 14.01 = 65.02 \text{ g/mol} \)
Mass: \( m = n \times M = 1.093 \text{ mol} \times 65.02 \text{ g/mol} \approx 71.0 \text{ g} \)

Answer:

\( \approx 71.0 \text{ g} \) (or more precisely, around 70.9 - 71.1 g depending on calculation precision)