Chất rắn không tan là Cu
=> mCu = 3,2 (g)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2<------------------------0,3
=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{3,2}{3,2+0,2.27}.100\%=37,21\%\\\%m_{Al}=\dfrac{0,2.27}{3,2+0,2.27}.100\%=62,79\%\end{matrix}\right.\)