E,m tham khảo nha :

\(\begin{array}{l}
a)\\
{H_2}S{O_4} + 2KOH \to {K_2}S{O_4} + 2{H_2}O\\
{m_{{H_2}S{O_4}}} = \dfrac{{200 \times 4,9}}{{100}} = 9,8g\\
{n_{{H_2}S{O_4}}} = \dfrac{{9,8}}{{98}} = 0,1mol\\
{n_{KOH}} = 2{n_{{H_2}S{O_4}}} = 0,2mol\\
{m_{KOH}} = 0,2 \times 56 = 11,2g\\
C{\% _{KOH}} = \dfrac{{11,2}}{{200}} \times 100\%  = 5,6\% \\
b)\\
{n_{{K_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,1mol\\
{m_{{K_2}S{O_4}}} = 0,1 \times 175 = 17,4g\\
C{\% _{{K_2}S{O_4}}} = \dfrac{{17,4}}{{200 + 200}} \times 100\%  = 4,35\% 
\end{array}\)

2KOH+H2SO4→K2SO4+2H2O

mH2SO4=200.4,9/100=9,8g

nH2SO4=9,8/98=0,1mol

→nKOH=0,2mol

mKOH=0,2.56=11,2g

C% KOH=11,2/200.100=5,6%

mdd spu=200+200=400g

mK2SO4=0,1.174=17,4g

C% K2SO4=17,4/400.100%=4,35%