If an automobile air bag has a volume of 11.7 L, what mass of NaN3 (in g) is required to fully inflate the air bag upon impact

33.95 grams of NaN3

Explanation:

Number of moles of NaN3 = mass (m) / MW = m/65 mole

I mole of NaN3 requires 22.4L air bag

m/65 mole of NaN3 required 11.7L

22.4m/65 = 11.7

22.4m = 65*11.7

22.4m = 760.5

m = 760.5/22.4 = 33.95grams of NaN3