\(n_{H_2SO_4}=\dfrac{98.5\%}{98}=0,05\left(mol\right)\\ PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4}=n_{CuO}=n_{H_2SO_4}=0,05\left(mol\right)\\ a,m_{CuO}=0,05.80=4\left(g\right)\\ b,m_{CuSO_4}=0,05.160=8\left(g\right)\\ m_{ddCuSO_4}=98+4=102\left(g\right)\\ C\%_{ddCuSO_4}=\dfrac{8}{102}.100\approx7,843\%\)