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

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

$4\times 27\mathrm{g} \mathrm{of} \mathrm{Al}\to 3\times 2\times 16\mathrm{g} \mathrm{of} {\mathrm{O}}_2(\mathrm{form} \mathrm{the} \mathrm{above} \mathrm{reaction})$

108g of Al $\to$ 96 g ${\mathrm{O}}_2$

27 g of Al $\to$ ?

108 × $X$ = 96 × 27

$\mathrm{X}=\frac{96\times 27}{108}$

=24g