Đáp án:

\(\begin{array}{l}
b)\\
{m_{{\rm{dd}}KOH}} = 112g\\
c)\\
C{\% _{{K_2}S{O_4}}} = 10,235\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{K_2}O + {H_2}O \to 2KOH\\
b)\\
{n_{{K_2}O}} = \dfrac{m}{M} = \dfrac{{18,8}}{{94}} = 0,2mol\\
{n_{KOH}} = 2{n_{{K_2}O}} = 0,4mol\\
{m_{KOH}} = n \times m = 0,4 \times 56 = 22,4g\\
{m_{{\rm{dd}}KOH}} = \dfrac{{22,4 \times 100}}{{20}} = 112g\\
c)\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = V \times d = 200 \times 1,14 = 228g\\
{m_{{\rm{dd}}spu}} = 112 + 228 = 340g\\
{n_{{K_2}S{O_4}}} = \dfrac{{{n_{KOH}}}}{2} = 0,2mol\\
{m_{{K_2}S{O_4}}} = n \times M = 0,2 \times 174 = 34,8g\\
C{\% _{{K_2}S{O_4}}} = \dfrac{{34,8}}{{340}} \times 100\%  = 10,235\% 
\end{array}\)