PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
Ta có: \(n_{H_2sO_4}=\dfrac{122,5\cdot40\%}{98}=0,5\left(mol\right)=n_{CuO}=n_{CuSO_4}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CuSO_4}=0,5\cdot160=80\left(g\right)\\m_{CuO}=0,5\cdot80=40\left(g\right)\end{matrix}\right.\)