Đáp án + Giải thích các bước giải:

$a\bigg)$

Đặt $n_{Zn}=x(mol);n_{Mg}=y(mol)$

$=>65x+24y=2,02(1)$

$n_{H_2}=\frac{1,12}{22,4}=0,05(mol)$

$Zn+2HCl\to ZnCl_2+H_2$

$Mg+2HCl\to MgCl_2+H_2$

Theo PT: $x+y=0,05(2)$

Từ $(1)(2)=>x=0,02(mol);y=0,03(mol)$

$=>\%m_{Zn}=\frac{0,02.65}{2,02}.100\%\approx 64,36\%$

$=>\%m_{Mg}=100-64,36=35,64\%$

$b\bigg)$

Theo PT: $n_{ZnCl_2}=n_{Zn}=0,02(mol);n_{MgCl_2}=n_{Mg}=0,03(mol)$

Ta có $m_{\rm dd\,sau\,pứ}=2,02+50-0,05.2=51,92(g)$

$\to \begin{cases} C\%_{ZnCl_2}=\dfrac{0,02.136}{51,92}.100\%\approx 5,24\%\\ C\%_{MgCl_2}=\dfrac{0,03.95}{51,92}.100\%\approx 5,49\% \end{cases}$

 `a)`

`Zn(x); Mg(y)` 

`n_{H_2} = {1,12}/{22,4} = 0,05(mol)`

`65x + 24y = 2,02(g)(1)`

Bte: `2x + 2y = 0,05 .2(2)`

`(1)(2) →x = 0,02(mol); y = 0,03(mol)`

`%m_{Zn} = {0,02.  65}/{2,02} .100 = 64,36%`

`%m_{Mg} = 100- 63,36= 35,64%`

`b)`

`mdd = 2,02 + 50 - 0,05 . 2 = 51,92(g)`

Bảo toàn `Mg → n_{MgCl_2} = 0,03(mol)`

Bảo toàn `Zn→ n_{ZnCl_2} = 0,02(mol)`

`C%_{ZnCl_2} = {0,02. 136}/{51,92} .100 = 5,24%`

`C_{MgCl_2} = {0,03.95}/{51,92} .100 = 5,49%`