What mass of sodium azide would be needed to liberate nitrogen?

Question

Please could you help me work out what mass would be needed, I got the answer as 84.5. But im unsure if its correct!

Equation: 2NaN3(s) ---------- 2Na(s) + 3N2(g)

What mass of sodium azide (2NaN3) would be needed to liberate 48dm3 of nitrogen? (Molar volume of gas is 24dm3mol-1)

Answer 1

For this question, we must assume the gas is at STP although it doesn't specify.

Technically, 1 mole of any gas at STP is 22.4 liters or 22.4 dm^3 but if you want to use 24, fine.

If you look at the balanced equation in reverse, you see for each mole of nitrogen gas created, it only take 2/3 of a mole of NaN3 (thus; for 3 moles of nitrogen you need (3x2)/3 or 2 moles of NaN3).

2 moles N2 x (2/3) x [23 + (3 x 14)]g/mol = 86.7g

or using 22.4L/mole for gas we get:

2.14 moleN2 x (2/3) x [23 + (3 x 14)]g/mol = 92.7g