\(a,PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(n_{CuO}=\dfrac{m}{M}=\dfrac{32}{80}=0,4\left(mol\right)\)
\(Theo.PTHH:n_{H_2SO_4}=n_{CuSO_4}=0,4\left(mol\right)\\ m_{H_2SO_4}=n.M=0,4.98=39,2\left(g\right)\\ b,m_{CuSO_4}=n.M=0,4.160=64\left(g\right)\)

