Sau phản ứng oxi hoá nhôm, dư nhôm nên tạo khí hidro.
$n_{O_2}=\dfrac{6,72}{22,4}=0,3(mol)$
$4Al+3O_2\xrightarrow{{t^o}} 2Al_2O_3$
$\to n_{Al\text{pứ}}=\dfrac{4}{3}n_{O_2}=0,4(mol)$
$n_{H_2}=\dfrac{13,44}{22,4}=0,6(mol)$
$Al_2O_3+6HCl\to 2AlCl_3+3H_2O$
$2Al+6HCl\to 2AlCl_3+3H_2$
$\to n_{Al\text{dư}}=\dfrac{2}{3}n_{H_2}=0,4(mol)$
Vậy $m_{Al}=27(0,4+0,4)=21,6g$