Đáp án:
\(\begin{array}{l}
{m_{{C_2}{H_5}OH}} = 20,7g\\
{V_{{H_2}}} = 23,94l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a,{C_2}{H_5}OH + K \to {C_2}{H_5}OK + \dfrac{1}{2}{H_2}\\
b,\\
{V_{{C_2}{H_5}OH}} = \dfrac{{{V_{{\rm{dd}}({C_2}{H_5}OH + {H_2}O)}} \times 46}}{{100}} = 25,875ml\\
{m_{{C_2}{H_5}OH}} = 25,875 \times 0,8 = 20,7g\\
c,\\
\end{array}\)
\(\begin{array}{l}
{C_2}{H_5}OH + K \to {C_2}{H_5}OK + \dfrac{1}{2}{H_2}\\
{H_2}O + K \to KOH + \dfrac{1}{2}{H_2}\\
{n_{{C_2}{H_5}OH}} = 0,45mol\\
{V_{{H_2}O}} = 56,25 - 25,875 = 30,375ml\\
\to {m_{{H_2}O}} = 1 \times 30,375 = 30,375g\\
\to {n_{{H_2}O}} = 1,6875mol\\
\to {n_{{H_2}}} = \dfrac{1}{2}({n_{{C_2}{H_5}OH}} + {n_{{H_2}O}}) = 1,06875mol\\
\to {V_{{H_2}}} = 23,94l
\end{array}\)
