a) $ 4Al+3O_{2}\rightarrow 2Al_{2}O_{3}(1)$ $ Al_{2}O_{3}+ 6HCl\rightarrow 2AlCl_{3}+3H_{2}O(2)$
b) $n_{Al}=0,1 => n_{O_{2}}=0,1*3/4=0,075(mol) => V_{O_{2}}=1,68(l)$
c) $n_{Al}=0,1=> n_{Al_{2}O_{3}}=0,1*2/4=0,05(mol)$.
Từ (2) => $n_{HCl}=6n_{Al_{2}O_{3}}=0,05*6=0,3(mol)=> m_{HCl}=0,3*36,5=10,95(g)$