\(n_{HCl}=0,05.2=0,1\left(mol\right)\)
Cu không tác dụng với dd HCl.
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(n_{Mg}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
\(m_{Mg}=0,05.24=1,2\left(g\right)\)
\(\%m_{Mg}=\dfrac{1,2}{4}.100\%=30\%\\ \%m_{Cu}=100\%-30\%=70\%\)