CuO + 2HCl -> CuCl2 + H2O
x -> ....................x (mol)
Al2O3 + 6HCl -> 2AlCl3 + 3H2O
y -> .....................2y (mol)
Giải hệ \(\left\{{}\begin{matrix}80a+102y=104,5\\135a+2y.133,5=247,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,35\\y=0,75\end{matrix}\right.\)
=> mCuO = 80 . 0,35 =28(g)
=> mAl2O3 = 104,5 - 28 =76,5(g)