Đáp án + Giải thích các bước giải:

$a\bigg)$

Đặt $n_{Mg}=x(mol);n_{Zn}=y(mol)$

$\to 24x+65y=5,65(1)$

$Mg+2HCl\to MgCl_2+H_2$

$Zn+2HCl\to ZnCl_2+H_2$

Theo PT: $n_{MgCl_2}=x(mol);n_{ZnCl_2}=y(mol)$

$\to 95x+136y=16,3(2)$

Từ $(1)(2)\to x=0,1(mol);y=0,05(mol)$

$\to \%m_{Mg}=\frac{0,1.24}{5,65}.100\%\approx 42,48\%$

$\to \%m_{Zn}=100-42,48=57,52\%$

$b\bigg)$

Theo PT: $\sum n_{HCl(pứ)}=2n_{Mg}+2n_{Zn}=0,3(mol)$

$\to n_{HCl(dư)}=\frac{100.21,9\%}{36,5}-0,3=0,3(mol)$

Theo PT: $\sum n_{H_2}=\frac{1}{2}\sum n_{HCl(pứ)}=0,15(mol)$

$m_{dd\,sau\,pứ}=5,65+100-0,15.2=105,35(g)$

$\to\begin{cases} C\%_{MgCl_2}=\frac{0,1.95}{105,35}.100\%\approx 9,02\%\\ C\%_{ZnCl_2}=\frac{0,05.136}{105,35}.100\%\approx 6,54\% \\ C\%_{HCl(dư)}=\frac{0,3.36,5}{105,35}.100\%\approx 10,39\% \end{cases}$