Chemistry of an airbag (Gas laws)

Question

the balanced equation

2NaN3(s) = 2Na+3N2

the air bag which contain 50.0g of NaN3 is about 60 L

If 50.0 g of NaN3 decompose how many moles of N2(g) will be produced? How many grams of N2(g) will be produced?

also how many moles of Na will be formed?

Answer 1

2NaN₃(s) ==> 2Na + 3N₂ ... balanced equation

First, find moles of NaN₃ present:

50.0 g NaN₃ x 1 mol NaN₃/65.0 g = 0.769 moles NaN₃

Next, find moles of N₂ produced from 0.769 moles of NaN₃:

0.769 moles NaN₃ x 3 moles N₂/2 mol NaN₃ = 1.15 moles N₂ produced

Converting this to grams, we obtain... 1.15 moles N₂ x 28.0 g N₂/mol N₂ = 32.2 g N₂ produced

Finally, to find moles of Na produced, again we use the stoichiometry of the balanced equation:

0.769 moles NaN₃ x 2 mol Na/2 mol NaN₃ = 0.769 moles Na produced