Đáp án:
b) 51,6 g
c) 75%
Giải thích các bước giải:
a)
$\begin{gathered}
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2} \hfill \\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2} \hfill \\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O \hfill \\
\end{gathered} $
b) ${n_{C{H_3}COOH}} = {n_{NaOH}} = 0,2mol$
$\begin{gathered}
{n_{{H_2}}} = \dfrac{1}{2}{n_{{C_2}{H_5}OH}} + \dfrac{1}{2}{n_{C{H_3}COOH}} = 0,25 \hfill \\
\Rightarrow {n_{{C_2}{H_5}OH}} = 0,25.2 - 0,2 = 0,3mol \hfill \\
\end{gathered} $
${m_X} = 2.({m_{{C_2}{H_5}OH}} + {m_{C{H_3}COOH}}) = 2.(46.0,3 + 60.0,2) = 51,6g$
c) ${C_2}{H_5}OH + C{H_3}COOH \to C{H_3}COO{C_2}{H_5} + {H_2}O$
Do ${n_{{C_2}{H_5}OH}} > {n_{C{H_3}COOH}}$ ⇒ hiệu suất phản ứng tính theo ${C{H_3}COOH}$
$\begin{gathered}
{n_{este}} = \dfrac{{26,4}}{{88}} = 0,3mol \hfill \\
\Rightarrow {n_{C{H_3}COOHpu}} = {n_{este}} = 0,3mol \hfill \\
\Rightarrow H = \dfrac{{{n_{C{H_3}COOHpu}}}}{{{n_{C{H_3}COOHbd}}}}.100\% = \frac{{0,3}}{{0,4}}.100\% = 75\% \hfill \\
\end{gathered} $