$a.$

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

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

$n_{H_2}=\dfrac{0.8}{2}=0.4(mol)$

$2Al+6HCl \to 2AlCl_3+3H_2$

$x$                  `->`          $\dfrac{3}{2}a$

$Mg+2HCl \to MgCl_2+H_2$

$y$                `->`            $y$

$2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2$

$z$                  `->`          $\dfrac{3}{2}z$

$Mg+H_2SO_4 \to MgSO_4+H_2$

$t$                `->`            $t$

`->`$\dfrac{3}{2}x+\dfrac{3}{2}t+y+t=n_{H_2}$

`<=>`$\dfrac{3}{2}(x+t)+(y+t)=0.4$

`<=>`$\dfrac{3}{2}a+b=0.4$ $(2)$

Từ $(1),(2)$ suy ra: $a=0.15;b=0.175$

`=>`$\%m_{Al}=\dfrac{0.15 \times 27}{7.8} \times 100\%=51.92\%$

$\%m_{Mg}=100\%-51.92\%=48.08\%$

$b.$

$n_{HCl}=0.8(mol)$

$n_{H_2SO_4}=0.8 \times 0.5=0.4(mol)$

Giả sử chỉ tạo muối chloride.

Bảo toàn nguyên tố $Al$`->`$n_{AlCl_3}=n_{Al}=0.15(mol)$

Bảo toàn nguyên tố $Cl$`->`$n_{MgCl_2}=\dfrac{n_{HCl}-3n_{AlCl_3}}{2}=0.175(mol)$

$m_{muối,1}=m_{MgCl_2}+m_{AlCl_3}=36.65(gam)$

Giả sử chỉ tạo muối sulfate.

Bảo toàn nguyên tố $Al$`->`$n_{Al_2(SO_4)_3}=\dfrac{n_{Al}}{2}=0.075(mol)$

Bảo toàn nguyên tố $S$`->`$n_{MgSO_4}=n_{H_2SO_4}-3n_{Al_2(SO_4)_3}=0.175(mol)$

$m_{muối,2}=m_{MgSO_4}+m_{Al_2(SO_4)_3}=46.65(gam)$

Vậy khối lượng muối sinh ra nằm trong khoảng $m_{muối,1}<m_{muối}<m_{muối,2}$

Hay $36.65<m_{muối}<46.65$

`a)`

`Hh:Al:x(mol);Mg:y(mol)`

`24x+27y=7,8(1)`

`n_{HCl}=0,7(mol);n_{H_2SO_4}=0,4(mol)`

`2Al+6HCl->2AlCl_3+3H_2`

`Mg+2HCl->MgCl_2+H_2`

`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`

`Mg+H_2SO_4->MgSO_4+H_2`

`->n_{H_2}=1,5x+y={0,8}/2=0,4(2)`

`(1)(2)->x=0,2;y=0,1`

`\%m_{Al}={0,2.27}/{7,8}.100\%=69,23\%`

`\%m_{Mg}=100-69,23=30,77\%`

`b)`

TH1: Chỉ có HCl pu

`->n_{AlCl_3}=x=0,2(mol);n_{MgCl_2}=y=0,1(mol)`

`->m_{m u o i\ min}=0,2.133,5+0,1.95=36,2(g)`

TH2: Chỉ có `H_2SO_4` pu

`->n_{Al_2(SO_4)_3}=1/2x=0,1(mol);n_{MgSO_4}=y=0,1(mol)`

`->m_{m u o i\ max}=0,1.342+0,1.120=46,2(g)`

Vậy `36,2\le m_{m u o i}\le 46,2`