Bạn tham khảo:
$2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2$
$n_{Al}=\frac{5,4}{27}=0,2(mol)$
$n_{H_2}=1,5.n_{Al}=1,5.0,2=0,3(mol)$
$a/$
$V_{H_2}=0,3.22,4=6,72(l)$
$b/$
$m_{Al_2(SO_4)_3}=0,1.342=34,2(g)$
$c/$
$n_{H_2SO_4}=1,5.n_{Al}=1,5.0,2=0,3(mol)$
$m_{ddH_2SO_4}=\frac{0,3.98.100}{9,8}300(g)$
$d/$
$2Al+6HCl \to 2AlCl_3+3H_2$
$n_{HCl}=0,5(mol)$
$\frac{0,2}{2}=0,1 > \frac{0,5}{6}$
$Al$ dư, $HCl$ hết
$m_{H_2}=0,25.2=0,5(g)$