Hòa tan hoàn toàn m gam hỗn hợp X gồm Al, Mg, Zn và Fe bằng dung dịch HCl dư, sau phản ứng thu được 2,688 lít khí H2 (đktc) và dung dịch chứa 13,5 gam

Cho $\begin{cases} Al: x(mol)\\ Mg: y(mol)\\ Zn: z(mol)\\ Fe: t(mol)\\\end{cases}$

`n_{H_2}=\frac{2,688}{22,4}=0,12(mol)`

Bảo toàn `H`: `=> n_{HCl}=2n_{H_2}=0,12.2=0,24(mol)`

BTKL:

`m=m_{\text{muối}}+m_{H_2}-m_{HCl}`

`=> m=13,5+0,12.2-0,24.36,5=4,98g`

BTe:

$\mathop{Fe}\limits^{0}\to \mathop{Fe}\limits^{+2}+2e$

$\mathop{Al}\limits^{0}\to \mathop{Al}\limits^{+3}+3e$

$\mathop{Zn}\limits^{0}\to \mathop{Zn}\limits^{+2}+2e$

$\mathop{Mg}\limits^{0}\to \mathop{Mg}\limits^{+2}+2e$

$\mathop{2H}\limits^{+1}+2e\to \mathop{H_2}\limits^{0}$

`n_{HCl}=3x+2y+2x+2t=0,24(mol)(1)`

Mặc khác:

`n_{Cl_2}=\frac{2,912}{22,4}=0,13(mol)`

BTe:

$\mathop{Fe}\limits^{0}\to \mathop{Fe}\limits^{+3}+3e$

$\mathop{Al}\limits^{0}\to \mathop{Al}\limits^{+3}+3e$

$\mathop{Zn}\limits^{0}\to \mathop{Zn}\limits^{+2}+2e$

$\mathop{Mg}\limits^{0}\to \mathop{Mg}\limits^{+2}+2e$

$\mathop{Cl_2}\limits^{0}+2e\to \mathop{2Cl}\limits^{-1}$

Ta có:` 3x+2y+2z+3t=2n_{Cl_2}`

`=>3x+2y+2x+3t=0,26(mol)(2)`

Lấy `(2)-(1)` ta được:

`t=0,02(mol)`

`=> %m_{Fe}=\frac{0,02.56.100%}{4,98}\approx 22,5%`