Em tham khảo nha :
\(\begin{array}{l}
a)\\
{H_2}S{O_4} + 2KOH \to {K_2}S{O_4} + 2{H_2}O\\
{m_{{H_2}S{O_4}}} = \dfrac{{200 \times 4,9}}{{100}} = 9,8g\\
{n_{{H_2}S{O_4}}} = \dfrac{{9,8}}{{98}} = 0,1mol\\
{n_{KOH}} = 2{n_{{H_2}S{O_4}}} = 0,2mol\\
{m_{KOH}} = 0,2 \times 56 = 11,2g\\
C{\% _{KOH}} = \dfrac{{11,2}}{{200}} \times 100\% = 5,6\% \\
b)\\
{n_{{K_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,1mol\\
{m_{{K_2}S{O_4}}} = 0,1 \times 174 = 17,4g\\
C{\% _{{K_2}S{O_4}}} = \dfrac{{17,4}}{{200 + 200}} \times 100\% = 4,35\%
\end{array}\)
