Đặt nFe=a nZn=b(mol)
Từ đó suy ra \(56a+65b=25,1\left(1\right)\)
\(n_{HCl}=0,8\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)
Lại suy ra được \(2a+2b=0,8\left(2\right)\)
Suy ra a=0,1 b=0,3 (mol)
\(V_{H_2}=\dfrac{0,8\cdot22,4}{2}=8,96\left(l\right)\\ m_{muoi}=m_{KL}+m_{Cl}=25,1+0,8\cdot35,5=53,5\left(g\right)\\ \%m_{Fe}=\dfrac{0,1\cdot56}{25,1}\cdot100\%\approx22,31\left(\%\right)\\ \%m_{Zn}=100\%-22,31\%=77,69\left(\%\right)\)