Ta có: mH2SO4=(200.19.6):100=39,2g

          →nH2SO4=39,2:98=0,4 mol

     PTHH:Cu(OH)2+H2SO4→CuSO4+2H2O  (1)

 Theo đề:  0.2mol    0.4mol 

           →H2SO4 dư,Cu(OH)2 hết

Theo PT(1):nCuSO4=0,2 mol→mCuSO4=0,2.160=32g

                  nH2SO4 dư=0,4-0,2=0.2 mol→mH2SO4=0,2.98=19,6g

⇒mCu(OH)2=0,2.98=19,6g

⇒mdd thu dc=200+19,6=219,6g

⇒  C%H2SO4 dư=(19,6:219,6).100%=8,93%

C%CuSO4=(32:219,6).100%=14,57%

Em tham khảo nha :

\(\begin{array}{l}
Cu{(OH)_2} + {H_2}S{O_4} \to CuS{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{200 \times 19,6\% }}{{98}} = 0,4mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,4}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{{H_2}S{O_4}d}} = 0,4 - 0,2 = 0,2mol\\
C{\% _{{H_2}S{O_4}d}} = \dfrac{{0,2 \times 98}}{{0,2 \times 98 + 200}} \times 100\%  = 8,93\% \\
{n_{CuS{O_4}}} = {n_{Cu{{(OH)}_2}}} = 0,2mol\\
{m_{CuS{O_4}}} = 0,2 \times 160 = 32g\\
C{\% _{CuS{O_4}}} = \dfrac{{32}}{{0,2 \times 98 + 200}} \times 100\%  = 14,6\% 
\end{array}\)