\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,15<----------------------0,15
\(m_{Fe}=0,15.56=8,4\left(g\right)\\ \Rightarrow\%m_{Fe}=\dfrac{8,4}{10,5}.100\%=80\%\\ \Rightarrow\%m_{Cu}=100\%-80\%=20\%\)