a, \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
b, Ta có: 80nCuO + 81nZnO = 12,1 (1)
Theo PT: \(n_{HCl}=2n_{CuO}+2n_{ZnO}=0,2.1,5=0,3\left(mol\right)\left(2\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{CuO}=0,05\left(mol\right)\\n_{ZnO}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,05.80}{12,1}.100\%\approx33,06\%\\\%m_{ZnO}\approx66,94\%\end{matrix}\right.\)