What weight of $\mathrm{Al}$ will be completely oxidized by $44.8$ liters of oxygen at STP?

Aluminum reacts with oxygen to form aluminum oxide. The balanced chemical equation is,

               $4\mathrm{Al}+3{\mathrm{O}}_2\to 2{\mathrm{Al}}_2{\mathrm{O}}_3$

4 moles of aluminum combined with 3 moles of oxygen.

$4\times 27 \mathrm{g}$ of $\mathrm{Al}$ combines with $3\times 22.4 \mathrm{L}$ of $O_2$

So, 44.8L of $O_2$ consumes $=\frac{44.8\times 4\times 27}{3\times 22.4}=72 \mathrm{g}$ of $\mathrm{Al}$