Đáp án:
$a)$
\[n_{H_2}=\dfrac{12,32}{22,4}=0,55(mol)\]
Gọi số mol Fe, Zn lần lượt là $a;b$
\[Fe+2HCl\to FeCl_2+H_2\]
\[Zn+2HCl\to ZnCl_2+H_2\]
Ta có phương trình:
\[\left\{\begin{matrix} 56a+65b=32,6 & \\ a+b=0,55& \end{matrix}\right.\] \[\to \left\{\begin{matrix} a=0,35(mol) & \\ b=0,2(mol)& \end{matrix}\right.\]
\[\to m_{Fe}=0,35\times 56=19,6(g)\]
\[\to \%m_{Fe}=\dfrac{19,6}{32,6}\times 100\%=60,12\%\]
\[\to \%m_{Zn}=100\%-60,12\%=39,88\%\]
$b)$ \[n_{Fe}=n_{FeCl_2}=0,35(mol)\]
\[\to m_{FeCl_2}=0,35\times 127=44,45(g)\]
\[n_{ZnCl_2}=n_{Zn}=0,2(mol)\]
\[\to m_{ZnCl_2}=0,2. 136=27,2(g)\]