Giải thích các bước giải:
\(\begin{array}{l}
a)2K + 2{H_2}O \to 2KOH + {H_2}\\
b)\\
{n_K} = 0,2mol\\
\to {n_{KOH}} = {n_K} = 0,2mol \to {m_{KOH}} = 11,2g\\
\to {n_{{H_2}}} = \dfrac{1}{2}{n_K} = 0,1mol \to {m_{{H_2}}} = 0,2g\\
{m_{{\rm{dd}}{H_2}O}} = 100g\\
\to {m_{{\rm{dd}}}} = {m_K} + {m_{{\rm{dd}}{H_2}O}} - {m_{{H_2}}} = 107,6g\\
\to C{\% _{KOH}} = \dfrac{{11,2}}{{107,6}} \times 100\% = 10,4\% \\
c)\\
{V_{{H_2}}} = 2,24l\\
d)\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{1}{2}{n_{KOH}} = 0,1mol\\
\to {V_{{H_2}S{O_4}}} = \dfrac{{0,1}}{1} = 0,1l
\end{array}\)
