Đáp án:

 $↓$

Giải thích các bước giải:

$a)\\Fe+H_2SO_4 \to FeSO_4+H_2\uparrow\\2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2 \uparrow\\\begin{cases}n_{Fe}=x(mol)\\n_{Al}=y(mol)\\\to 56x+27y=8,3\end{cases}\\\to n_{H_2}=\dfrac{5,6}{22,4}=x+\dfrac32y=0,25\\\begin{cases}x+\dfrac32y=0,25\\56x+27y=8,3\end{cases} \Leftrightarrow \begin{cases}x=0,1(mol)\\y=0,1(mol)\end{cases}\\\Rightarrow \%m_{Fe}=\dfrac{0,1 \times 56 \times 100}{8,3} \approx 67,47(\%)\\\Rightarrow \%m_{Al}=32,53(\%)\\b)\\m_{FeSO_4}=152 \times 0,1=15,2(g)\\m_{Al_2(SO_4)_3}=(0,1)/2 \times 342=17,1(g)\\\Rightarrow m_{\text{chất rắn khan}}=15,2+17,1=32,3(g)$

Đáp án:

`%m_{Fe}~~67,47%\ ; %m_{Al}=32,53%`

`m_{m uoi}=32,3(g)`

Giải thích các bước giải:

`Pt:Fe+H_2SO_4->FeSO_4+H_2`

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

`a)`

`n_{H_2}={5,6}/{22,4}=0,25(mol)`

`Goi` $\begin{cases} 56a+27b=8,3\\a+1,5b=0,25 \end{cases}$ `=>` $\begin{cases} a=0,1\\b=0,1 \end{cases}$

`%m_{Fe}={0,1.56}/{8,3}.100%~~67,47%`

`%m_{Al}=100%-67,47%=32,53%`

`b)`

`n_{FeSO_4}=n_{Fe}=0,1(mol)`

`n_{Al_2(SO_4)_3}=1/{2}n_{Fe}=0,05(mol)`

`m_{m uoi}=m_{FeSO_4}+m_{Al_2(SO_4)_3}`

`m_{m uoi}=0,1.152+0,05.342=32,3(g)`