Đáp án + Giải thích các bước giải:

$a\big)$

Đặt $n_{Mg}=x(mol);n_{Zn}=y(mol)$

$Mg+2HCl\to MgCl_2+H_2↑$

$Zn+2HCl\to ZnCl_2+H_2↑$

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

$\to 95x+136y=59,8(1)$

Khi đốt cháy: $n_{Mg}=2x(mol);n_{Zn}=2y(mol)$

$2Mg+O_2\xrightarrow{t^o}2MgO$

$2Zn+O_2\xrightarrow{t^o}2ZnO$

Bảo toàn KL:

$m_{KL}+m_{O_2}=m_{oxit}$

$\to m_{O_2}=2m+16-2m=16(g)$

Theo PT: $2x.\frac{1}{2}+2y.\frac{1}{2}=n_{O_2}=\frac{16}{32}=0,5(2)$

Từ $(1)(2)\to x=0,2(mol);y=0,3(mol)$

$\to \%m_{Mg}=\frac{0,2.24}{0,2.24+0,3.65}.100\%\approx 19,75\%$

$\to \%m_{Zn}=100\%-19,75\%=80,25\%$

$b\big)$

Theo PT: `\sum n_{H_2}=x+y=0,5(mol)`

$\to V=V_{H_2(đktc)}=0,5.22,4=11,2(l)$