a) \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
b) \(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)=n_{H_2SO_4}=n_{CuSO_4}\)
\(m_{ddH_2SO_4}=\dfrac{0,04.98}{4,9\%}=80\%\)
\(m_{ddsaupu}=3,2+80=83,2\left(g\right)\)
=> \(C\%_{CuSO_4}=\dfrac{0,04.160}{83,2}.100=7,69\%\)