a,
Hỗn hợp: $\left\{\begin{matrix}Al:x\ (mol)\\Mg:y\ (mol)\end{matrix}\right.$
`=>27x+24y=7,8\ \ (1)`
`2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2`
`Mg+H_2SO_4\to MgSO_4+H_2`
`n_{H_2}=\frac{8,96}{22,4}=0,4\ (mol)`
`=>1,5x+y=0,4\ \ (2)`
Giải hệ `(1)`, `(2)` ta được: $\begin{cases}x=0,2\ (mol)\\y=0,1\ (mol)\end{cases}$
`%m_{Al}=\frac{0,2.27}{7,8}.100%=69,23%`
`%m_{Mg}=100%-69,23%=30,68%`
b,
Theo PT: $\left\{\begin{matrix}n_{MgSO_4}=n_{Mg}=0,1\ (mol)\\n_{Al_2(SO_4)_3}=\dfrac12n_{Al}=\dfrac12.0,2=0,1\ (mol)\end{matrix}\right.$
`=>`$\left\{\begin{matrix}m_{MgSO_4}=0,1.102=12\ (g)\\m_{Al_2(SO_4)_3}=0,1.342=34,2\ (g)\end{matrix}\right.$
$m_{dd\ spứ}=m_{hh}+m_{dd\ H_2SO_4}-m_{H_2}$
$=7,8+200-0,4.2=207\ (g)$
`=>`$\left\{\begin{matrix}{C\%}_{dd\ MgSO_4}=\dfrac{12}{207}.100\%=5,8\%\\{C\%}_{dd\ Al_2(SO_4)_3}=\dfrac{34,2}{207}.100\%=16,5\%\end{matrix}\right.$
