27 g of Al will react completely with how many grams of oxygen ?

$\begin{array}{l}4\mathrm{Al}+3{\mathrm{O}}_2\to 2{\mathrm{Al}}_2{\mathrm{O}}_3\\ \mathrm{At}.\mathrm{Wt}.\mathrm{of}\mathrm{Al}=27\\ \mathrm{Thus}4\times 27\mathrm{g}\mathrm{of}\mathrm{Al}\mathrm{reacts}\mathrm{with}3\times 32\mathrm{g} \mathrm{ofoxygen}\\ \therefore \mathrm{mass} \mathrm{ofoxygen} \mathrm{that}\mathrm{reacts}\mathrm{with}27\mathrm{g}\mathrm{of}\mathrm{Al}=\frac{3\times 32}{4\times 27}\times 27=24\mathrm{g}.\end{array}$