Giải thích các bước giải:
\(\begin{array}{l}
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{KOH}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = 0,15mol\\
\to \dfrac{{{n_{KOH}}}}{2} > \dfrac{{{n_{{H_2}S{O_4}}}}}{1} \to {n_{{H_2}S{O_4}}}du\\
a)\\
{n_{{K_2}S{O_4}}} = \dfrac{1}{2}{n_{KOH}} = 0,1mol\\
\to {m_{{K_2}S{O_4}}} = 17,4g\\
\to {n_{{H_2}S{O_4}(pt)}} = \dfrac{1}{2}{n_{KOH}} = 0,1mol\\
\to {n_{{H_2}S{O_4}(dư)}} = 0,05mol\\
\to {m_{{H_2}S{O_4}(dư)}} = 4,9g\\
b)\\
C{M_{{K_2}S{O_4}}} = \dfrac{{0,1}}{{0,2 + 0,3}} = 0,2M\\
C{M_{{H_2}S{O_4}(dư)}} = \dfrac{{0,05}}{{0,2 + 0,3}} = 0,1M
\end{array}\)