$\Delta m = m_{KL} - m_{H_2}$

`<=> 7 = 7,8- 2n_{H_2}` 

`→ n_{H_2}=0,4(mol)` 

Bảo toàn H: 

`n_{HCl}= 2n_{H_2}= 0,8(mol)` 

`→ n_{Cl}= n_{HCl}= 0,8(mol)` 

`m_{text{muối}}= m_{KL}+ m_{Cl}`

`= 7,8 + 0,8. 35,5=36,2(g)`

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

`=>`$24a+27b=7.8$ $(1)$

Bảo toàn $e:$

$2a+3b=2n_{H_2}$

`=>`$n_{H_2}=a+1.5b(mol)$

$m_{dd,tăng}=m_{thêm}-m_{khí}=m_{Mg}+m_{Al}-m_{H_2}$

`->`$24a+27b-2(a+1.5b)=7$

`->`$22a+24b=7$ $(2)$

Từ $(1)$ và $(2)$ suy ra: $a=0.1;b=0.2$

Bảo toàn  nguyên  tố $Mg:$

$n_{MgCl_2}=n_{Mg}=0.1(mol)$

Bảo toàn  nguyên  tố $Al:$

$n_{AlCl_3}=n_{Al}=0.2(mol)$

`=>`$m_{muối}=m_{MgCl_2}+m_{AlCl_3}=36.2(gam)$