$\rm n_{Al_2O_3} = \dfrac{40,8}{102} = 0,4$ mol

Phản ứng xảy ra:

$\rm 4Al + 3O_2 \xrightarrow{t^\circ} 2Al_2O_3$

Theo PTHH:

$\rm n_{Al} = 2n_{Al_2O_3} = 2 \times 0,4 = 0,8$ mol

$\rm \to a = m_{Al} = 0,8 \times 27 = 21,6g$

Theo PTHH:

$\rm n_{O_2} = \dfrac{3}{2}n_{Al_2O_3} = \dfrac{3}{2} \times 0,4 = 0,6$ mol

$\rm \to b = V_{O_2} = 0,6 \times 24,79 = 14,874(l)$