\(a,\) Đặt \(n_{Zn}=x(mol);n_{Fe}=y(mol)\Rightarrow 65x+56y=16,1-4=12,1(1)\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ Fe+2HCl\to FeCl_2+H_2\\ b,x+y=0,2(2)\\ (1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Zn}=\dfrac{65.0,1}{16,1}.100\%=40,37\%\\ \%_{Fe}=\dfrac{56.0,1}{16,1}.100\%=34,78\%\\ \%_{Cu}=100\%-40,37\%-34,78\%=24,85\%\)