Air bag reaction calculation: gas volume and stoichiometry

Question

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?

Answer 1

Both of the answer keys are incorrect, it seems, and the OP's are correct. We can combine the two equations (in a way that no Na is left over) to make the calculations a bit easier:

$$\ce{10 NaN3 + 2 KNO3 -> 16 N2 + K2O + 5 Na2O}$$

For a), the mol-ratio of the two reactants is 5:1. Multiplying by the molar masses, the mass-ratio is 3.22:1 or 0.311:1 (using the same 3 significant figures as the OP).

To make 16 moles of $\ce{N2}$, you would need about 852 g of reactants at stoichiometric mixture. This would be a volume of 16 * 22.4 L. = 358.4 L. You only need 60 L though, so about one sixth of that. This comes out to 143 g of reactant mixture.