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I have come across the following problem:

One method used to inflate air bags in cars is to use nitrogen produced chemically from the decomposition of sodium azide. The sodium formed reacts with potassium nitrate to give more nitrogen.

$$\ce{2NaN3(s) -> 2Na(s) + 3N2(g)}$$

$$\ce{10Na(s) + 2KNO3(s) -> K2O(s) + 5Na2O(s) + N2(g)}$$

a) In what ratio by mass must the sodium azide and potassium nitrate be mixed in order that no metallic sodium remains after the reaction?

b) Calculate the total mass of the solid mixture needed to inflate a $60.0\ \mathrm{dm^3}$ air bag at room temperature and atmospheric pressure.

I have a solution to part a), which is 1:0.311, but this differs from the given answer (1:3.11). Here is my solution:

For every 10 moles of $\ce{NaN3}$, 2 moles of $\ce{KNO3}$ is needed i.e. ratio in moles is 5:1. Mass of $5\ \mathrm{mol}$ of $\ce{NaN3}$ is $5 \times 65.0 = 325$ and mass of 1 mol of $\ce{KNO3}$ is 101.1. So, ratio = 325:101.1 = 1:0.311. Is this incorrect?

I also have a solution to part b), but again this differs from the given answer $(40.9\ \mathrm g)$. Here is my solution: Using ideal gas law to find how many moles of $\ce{N2}$ needed to inflate air bag: $n=\frac{PV}{RT}=\frac{101000\times 0.06}{8.31\times293}=2.49\ \mathrm{mol}$ If the first equation is rewritten as $\ce{10NaN3(s) -> 10Na(s) + 15N2(g)}$, then there are $\frac{15}{16}\times2.49=2.33$ moles of $\ce{N2}$ from the decomposition of sodium azide, and $\frac{1}{16}\times2.19=0.156$ moles of $\ce{N2}$ from the second reaction. Using these values gives $101\ \mathrm g$ of $\ce{NaN3}$ and $31.5\ \mathrm g$ of $\ce{KNO3}$, which clearly does not give the answer. What's incorrect here?

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  • $\begingroup$ I suspect you're using the wrong gas equation. There is a much simpler one relating the number of moles of a gas directly to the volume. $\endgroup$ – Beerhunter Nov 11 '15 at 19:36
  • $\begingroup$ @Beerhunter, I assume your referring to the fact that 1 mole of gas always has a volume of 24 decimetres cubed at rtp, but this gives the same value as mine for the number of moles of nitrogen . $\endgroup$ – Ambler Nov 11 '15 at 20:15

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