\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,1 0,1
\(m_{Zn}=0,1.65=6,5\left(g\right)\\
\%m_{Zn}=\dfrac{6,5}{10,5}.100\%=61,9\%\\
\%m_{Cu}=100\%-61,9\%=38,1\%\)