Volume of nitrogen gas produced during taking off


> 
> The fuel in the buster rockets of the Space Shuttle is constituted of a mixture of ammonium perchlorate, $\ce{NH4ClO4}$, and aluminum powder. One of the reactions taking place during taking off is given by
>  $$\ce{6 NH4ClO4(s) + 10 Al(s) -> 5 Al2O3(s) + 3 N2(g) + 6 HCl(g) + 9 H2O(g)}$$
>  The indication (s) and (g) signify solid and gaseous phase. Suppose that these rockets are loaded with $\pu{11.75 ton}$ of ammonium perchlorate and $\pu{2.70 ton}$ of aluminum.
> 
> 
> Calculate the volume of nitrogen gas produced during taking off under normal
>  conditions ($\pu{0^\circ C}$ and atmospheric pressure, so that $\pu{1 mol}$ corresponds to $\pu{22.4 L}$ of gas).
> 
> 
> 


I got to the point where it's $\pu{10^5 mol}$ for each of the compounds.
How do you get to the answer $\pu{6.72e5 L}$?




Your next step is to identify the **limiting reagent**.


Observe that six moles of ammonium perchlorate react with ten moles of aluminum. So by taking ratios you find that:


1) The $10^5$ moles of ammonium perchlorate you started with react with approximately $1.7×10^5$ moles of aluminum.


2) The $10^5$ moles of aluminum you started with react with $6×10^4$ moles of ammonium perchlorate.


Now there is a problem with (1) -- you have only $1.0×10^5$ moles of aluminum available so not all of the ammonium perchlorate can react as indicated in (1). So we now look at (2) -- it indicates that all of the aluminum can react and you still have $4×10^4$ moles of ammonium perchlorate remaining.


Thus we can realize condition (2) rather than (1) because the amount of reaction is limited by the aluminum you have available. Thus **aluminum is the limiting reagent**.


Now that you know the limiting reagent you can infer how much product is formed. With $10^5$ moles of aluminum all being consumed, you should be able to see that $3×10^4$ moles of nitrogen molecules are formed as gas. The volume under normal conditions should then follow.


 3