\(n_{ZnCl_2}=\dfrac{100.10\%}{136}=\dfrac{5}{68}\left(mol\right)\)
\(Mg+ZnCl_2\rightarrow MgCl_2+Zn\)
\(\dfrac{5}{68}\) \(\dfrac{5}{68}\) ( mol )
\(m_{Mg}=\dfrac{5}{68}.24=1,76\left(g\right)\)
\(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{1,76}{100}.100=1,76\%\\\%m_{Cu}=100\%-1,76\%=98,24\%\end{matrix}\right.\)