\(n_{HCl}=2\cdot0,1=0,2\left(mol\right)\\ PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\\ \Rightarrow n_{CuCl_2}=n_{CuO}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{CuCl_2}=0,1\cdot135=13,5\left(g\right)\\m_{CuO}=0,1\cdot80=8\left(g\right)\end{matrix}\right.\)