Question

Automobile air bags inflate following a serious impact. Impact triggers the following chemical reaction:
2NaN3 (s)→2Na(s)+3N2 (g)
If an automobile air bag has a volume of 11.8 L, what mass of NaN3 (in grams) is required to fully inflate an air bag upon impact if the pressure is 956 mmHg and the temperature is 21 °C?

Answer 1

Given:

P = 956 mm Hg

= (956/760) atm

= 1.2579 atm

V = 11.8 mL

= (11.8/1000) L

= 0.0118 L

T = 21.0 oC

= (21.0+273) K

= 294 K

find number of moles using:

P * V = n*R*T

1.2579 atm * 0.0118 L = n * 0.08206 atm.L/mol.K * 294 K

n = 6.152*10^-4 mol

This is mol of N2 produced.

From given balanced reaction,

mol of NaN3 reacted = (2/3)*mol of N2 produced

= (2/3)*6.152*10^-4 mol

= 4.101*10^-4 mol

Molar mass of NaN3,

MM = 1*MM(Na) + 3*MM(N)

= 1*22.99 + 3*14.01

= 65.02 g/mol

use:

mass of NaN3,

m = number of mol * molar mass

= 4.101*10^-4 mol * 65.02 g/mol

= 2.666*10^-2 g

Answer: 2.67*10^-2 g