Question

2nan₃(s) → 2na(s) + 3n₂(g)
6na(s) + fe₂o₃(s) → 3na₂o(s) + 2fe

1. assume that 65.1 l of n₂ gas are needed to inflate an airbag to the proper size. how many grams of nan₃ must be included in the gas generator to generate this amount of n₂? (hint: the density of n₂ gas at this temperature is about 0.916 g/l).

Explanation:

Step1: Calculate mass of N₂

Mass = density × volume = $0.916\ \text{g/L} × 65.1\ \text{L}$
$= 59.6316\ \text{g}$

Step2: Find moles of N₂

Molar mass of N₂ = $2×14.01 = 28.02\ \text{g/mol}$
Moles = $\frac{\text{mass}}{\text{molar mass}} = \frac{59.6316\ \text{g}}{28.02\ \text{g/mol}}$
$≈ 2.128\ \text{mol}$

Step3: Relate moles of N₂ to NaN₃

From reaction: $2\ \text{mol NaN}_3 → 3\ \text{mol N}_2$
Moles of NaN₃ = $2.128\ \text{mol N}_2 × \frac{2\ \text{mol NaN}_3}{3\ \text{mol N}_2}$
$≈ 1.419\ \text{mol}$

Step4: Calculate mass of NaN₃

Molar mass of NaN₃ = $23.0 + 3×14.01 = 65.03\ \text{g/mol}$
Mass = moles × molar mass = $1.419\ \text{mol} × 65.03\ \text{g/mol}$
$≈ 92.3\ \text{g}$

Answer:

92.3 g