$4)$

$b)$

$n_{H_2}=0.33(mol)$

Bảo toàn nguyên tố $H:$

$n_{HCl}=2n_{H_2}=0.66(mol)$

$V=\dfrac{0.66}{2}=0.33(L)$

$a)$

$n_{Mg}=a(mol);n_{Al}=b(mol)$

`->`$24a+27b=6.84$ $(1)$

Bảo toàn $e:$

$2a+3b=2n_{H_2}=0.66$ $(2)$

Từ $(1),(2)$ suy ra : $a=0.15;b=0.12$

`=>`$m_{Mg}=3.6(gam);m_{Al}=3.24(gam)$

$\%m_{Mg}=52.63\%||\%m_{Al}=47.37$

$c)$

Bảo  toàn nguên tố $Mg:$

$n_{Mg}=n_{MgCl_2}=0.15(mol)$

Bảo  toàn nguên tố $Al:$

$n_{Al}=n_{AlCl_3}=0.12(mol)$

$C_M(MgCl_2)=\dfrac{0.15}{0.33}=\dfrac{5}{11}(M)$

$C_M(AlCl_3)=\dfrac{0.12}{0.33}=\dfrac{4}{11}(M)$