Sodium azide to Nitrogen ratio in airbags
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.


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