CuO + 2HCl \(\rightarrow\)CuCl2 + H2O (1)
Fe2O3 + 6HCl \(\rightarrow\)2FeCl3 + 3H2O (2)
nHCl=0,2.3,5=0,7(mol)
Đặt nCuO=a
nFe2O3=b
Ta có hệ pt:
\(\left\{{}\begin{matrix}80a+160b=20\\2a+6b=0,7\end{matrix}\right.\)
a=0,05;b=0,1
mCuO=80.0,05=4(g)
% Cu=\(\dfrac{4}{20}.100\%=20\%\)
% Fe=100-20=80%