Volume of nitrogen gas produced during taking off
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.