$n_{Al}=\frac{5,4}{27}=0,2(mol)$
$n_{Mg}=\frac{8,4}{24}=0,35(mol)$
$2Al+6HCl→2AlCl_3+3H_2↑$ (1)
-Theo phương trình $(1)$: $n_{HCl}=3.n_{Al}=3.0,2=0,6(mol)$
$Mg+2HCl→MgCl_2+H_2↑$ (2)
-Theo phương trình $(2)$: $n_{HCl}=2.n_{Mg}=2.0,35=0,7(mol)$
$⇒∑n_{HCl}=0,6+0,7=1,3(mol)$
$⇒CM_{HCl}=\frac{1,3}{2}=0,65(M)$