Giải thích các bước giải:
\(\begin{array}{l}
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{{K_2}S{O_4}}} = 0,05mol\\
a)\\
{n_{{H_2}S{O_4}}} = {n_{{K_2}S{O_4}}} = 0,05mol \to {m_{{H_2}S{O_4}}} = 4,9g\\
\to C{\% _{{H_2}S{O_4}}} = \dfrac{{4,9}}{{200}} \times 100\% = 2,45\% \\
b)\\
{n_{KOH}} = 2{n_{{K_2}S{O_4}}} = 0,1mol \to {m_{KOH}} = 5,6g\\
\to {m_{{\rm{dd}}KOH}} = \dfrac{{5,6}}{{10\% }} \times 100\% = 56g\\
c)\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}KOH}} + {m_{{\rm{dd}}{{\rm{H}}_2}S{O_4}}} = 256g\\
\to C{\% _{{K_2}S{O_4}}} = \dfrac{{8,7}}{{256}} \times 100\% = 3,4\%
\end{array}\)