ZnO + 2HCl -> ZnCl2 + H2O (1)
CuO + 2HCl -> CuCl2 + H2O (2)
nHCl=0,3(mol)
Đặt nZnO=a
nCuO=b
Ta có hệ:
\(\left\{{}\begin{matrix}m_{hh}=81a+80b=12,1\\n_{HCl}=2a+2b=0,3\end{matrix}\right.\)
=>a=0,1;b=0,05
mCuO=80.0,05=4(g)
%mCuO=\(\dfrac{4}{12,1}.100\%=33\%\)
%mZnO=100-33=67%