Giải thích các bước giải:
\(\begin{array}{l}
{K_2}O + {H_2}O \to 2KOH\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
a)\\
{n_{{K_2}O}} = 0,004mol\\
\to {n_{KOH}} = 2{n_{{K_2}O}} = 0,008mol\\
\to C{M_{KOH}} = \dfrac{{0,008}}{{0,2}} = 0,04M\\
b)\\
{n_{{K_2}S{O_4}}} = \dfrac{1}{2}{n_{KOH}} = 0,004mol\\
\to {m_{{K_2}S{O_4}}} = 0,696g\\
c)\\
{n_{{H_2}S{O_4}}} = \dfrac{1}{2}{n_{KOH}} = 0,004mol\\
\to {m_{{H_2}S{O_4}}} = 0,392g\\
\to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,392}}{{20\% }} \times 100\% = 1,96g\\
\to {V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{1,96}}{{1,14}} = 1,719ml
\end{array}\)