\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.1..................................0.15\)
\(m_{Al}=0.1\cdot27=2.7\left(g\right)\)
\(m_{Cu}=10-2.7=7.3\left(g\right)\)
\(\%m_{Al}=\dfrac{2.7}{10}\cdot100\%=27\%\)
\(\%m_{Cu}=73\%\)