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.

$\ce{2 NaN3 -&gt; 2 Na + 3 N_2}$
$\ce{10 Na + 2 KNO3 -&gt; K2O + 5 Na2O + N2}$
$\ce{K2O + Na2O + 2 SiO2 -&gt; K2O3Si + Na2O3Si}$

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)
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.


  
  $\ce{NaN3 -&gt; Na + 3/2 N_2}$
  $\ce{Na + 1/5 KNO3 -&gt; 1/10K2O + 1/2 Na2O + 1/10 N2}$