The weight of oxygen required to completely react with 27g of Al grams.

 

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

Mass of 1 mole of $Al=27 g$

Since, 2 moles of $Al$ required $\frac{3}{2}$ moles of oxygen

Therefore, 1 mole of $Al$ required $\frac{3}{4}$ moles of oxygen $=\frac{3}{4}\times 32=24 g$ of oxygen. 

Hence, mass of oxygen required to completely react with $27 g$ of $Al$ is 24 grams.