Question

automobile airbags inflate following a serious impact. the impact triggers the chemical reaction: 2nan₃(s) → 2na(s) + 3n₂(g) part a if an automobile airbag has a volume of 12.4 l, how much nan₃ in grams is required to fully inflate the airbag upon impact? assume stp conditions. express your answer in grams to three significant figures. mass = g

Explanation:

Step1: Find moles of \(N_2\) at STP

At STP (Standard Temperature and Pressure, \(0^\circ\text{C}\) and \(1\) atm), 1 mole of any ideal gas occupies \(22.4\) L. So, moles of \(N_2\), \(n_{N_2}=\frac{V}{22.4\,\text{L/mol}}=\frac{12.4\,\text{L}}{22.4\,\text{L/mol}}\approx0.5536\,\text{mol}\).

Step2: Relate moles of \(N_2\) to \(NaN_3\) via reaction

The reaction is \(2NaN_3(s)
ightarrow2Na(s)+3N_2(g)\). From the stoichiometry, \(3\) moles of \(N_2\) are produced from \(2\) moles of \(NaN_3\). So, moles of \(NaN_3\), \(n_{NaN_3}=\frac{2}{3}\times n_{N_2}=\frac{2}{3}\times0.5536\,\text{mol}\approx0.3691\,\text{mol}\).

Step3: Calculate molar mass of \(NaN_3\)

Molar mass of \(NaN_3\): \(Na = 22.99\,\text{g/mol}\), \(N = 14.01\,\text{g/mol}\). So, \(M_{NaN_3}=22.99 + 3\times14.01=22.99 + 42.03 = 65.02\,\text{g/mol}\).

Step4: Find mass of \(NaN_3\)

Mass \(m = n\times M\), so \(m_{NaN_3}=0.3691\,\text{mol}\times65.02\,\text{g/mol}\approx23.99\,\text{g}\approx24.0\,\text{g}\) (to three significant figures).

Answer:

\(24.0\)