# Sodium azide to Nitrogen ratio in airbags

I've been given this question and I don't know how to solve it.

Airbags are designed to protect the occupants of a car during a collision. They use the decomposition of sodium azide, $\ce{NaN3}$. Three chemical reactions occur rapidly after impact and produce nitrogen gas ($\ce{N2}$) to fill the airbag.

The equations summarise the overall process that takes place.

1. $\ce{2 NaN3 -> 2 Na + 3 N_2}$
2. $\ce{10 Na + 2 KNO3 -> K2O + 5 Na2O + N2}$
3. $\ce{K2O + Na2O + 2 SiO2 -> K2O3Si + Na2O3Si}$

What is the ratio between the number of sodium azide ($\ce{NaN3}$) molecules used and the number of nitrogen ($\ce{N2}$) molecules produced in the overall process?

I initially just used the first equation to get a ratio of 2:3, but obviously that's not right. I have no clue where to start. How do you do this? (How this can be considered a Year 10 question baffles me)

• It baffles me also. You need also to consider the second equation, which produces N2 gas as well. The first equation generates Na, and this is consumed in the second equation. – long Apr 25 '16 at 4:20

Assuming you have an excess of $$\ce{KNO3}$$, the ratio of $$\ce{NaN3}$$ consumed to $$\ce{N2}$$ produced would be 5:8.
Let's renormalize the first two equations, dividing the first through by 2, and the second through by 10. By doing this we can see, directly, that in the overall reaction, 3/2 + 1/10 = 8/5 = 1.6 molecules of $$\ce{N_2}$$ are produced for every molecule of $$\ce{NaN3}$$ consumed. If you prefer an integer ratio, it's 1:(8/5) = 5:8.
1. $$\ce{NaN3 -> Na + 3/2 N_2}$$
2. $$\ce{Na + 1/5 KNO3 -> 1/10K2O + 1/2 Na2O + 1/10 N2}$$
The initial ratio 2:3 is correct. $$\ce{2NaN3 = 2 Na}$$ molecules as well as $$\ce{6 N}$$ molecules. The products of the reaction, $$\ce{2Na}$$ is equal to two $$\ce{Na}$$ molecules and $$\ce{3N2}$$ is equal to $$\ce{6 N}$$ molecules. So as you can see the equation is perfectly balanced. Hope this helps.