a, PT: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
b, Ta có: \(n_{AlCl_3}=\dfrac{26,7}{133,5}=0,2\left(mol\right)\)
Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{AlCl_3}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{0,1.102}{15}.100\%=68\%\\\%m_{Cu}=32\%\end{matrix}\right.\)