Đáp án:
\(\begin{array}{l}
1.\\
{V_{{H_2}}} = 11,2l\\
C{\% _{KOH}} = 5,6\% \\
2.\\
y = {m_{{K_2}O}} = 202,1g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1.\\
K + {H_2}O \to KOH + \frac{1}{2}{H_2}\\
{n_{{H_2}}} = \dfrac{1}{2}{n_K} = 0,5mol\\
\to {V_{{H_2}}} = 11,2l\\
{n_{KOH}} = {n_K} = 1mol \to {m_{KOH}} = 56g\\
\to C{\% _{KOH}} = \dfrac{{56}}{{1000}} \times 100\% = 5,6\% \\
2.\\
{K_2}O + {H_2}O \to 2KOH\\
\left\{ \begin{array}{l}
C{\% _{KOH}} = 5,6\% \to {m_{KOH}} = 56\\
C{\% _{KOH}} = 24\% \to {m_{KOH}} = 240
\end{array} \right.\\
\to {n_{KOH}} = 4,3mol\\
\to {n_{{K_2}O}} = \dfrac{1}{2}{n_{KOH}} = 2,15mol\\
\to {m_{{K_2}O}} = 202,1g
\end{array}\)