\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(0.15.....0.3.......................0.15\)
\(m_{Mg}=0.15\cdot24=3.6\left(g\right)\)
\(m_{Cu}=10-3.6=6.4\left(g\right)\)
\(\%Mg=\dfrac{3.6}{10}\cdot100\%36\%\)
\(\%Cu=64\%\)
\(V_{dd_{HCl}}=\dfrac{0.3}{2}=0.15\left(l\right)\)