\(\begin{array}{l}
250ml=0,25l\\
n_{HCl}=0,25.2=0,5(mol)\\
Dat\,n_{CuO}=x(mol);n_{Fe_2O_3}=y(mol)\\
\to 80x+160y=16(1)\\
CuO+2HCl\to CuCl_2+H_2O\\
Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\
Theo\,PT:\,2x+6y=0,5(2)\\
(1)(2)\to x=0,1(mol);y=0,05(mol)\\
\to m_{CuO}=0,1.80=8(g);m_{Fe_2O_3}=16-8=8(g)
\end{array}\)