a) \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
b) \(n_{AlCl_3}=\dfrac{26,7}{133,5}=0,2\left(mol\right)\)
PTHH: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
0,1<-----------------0,2
=> \(\%Al_2O_3=\dfrac{0,1.102}{15}.100\%=68\%\Rightarrow\%Cu=100\%-68\%=32\%\)