Đáp án:
b) 1,66 g
Giải thích các bước giải:
a) $\begin{gathered}
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O \hfill \\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2} \hfill \\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2} \hfill \\
\end{gathered} $
b) $\begin{gathered}
{n_{C{H_3}COOH}} = {n_{NaOH}} = 0,2.0,1 = 0,02mol \hfill \\
{n_{{H_2}}} = 0,06mol \hfill \\
\end{gathered} $
Ta có: $\begin{gathered}
{n_{C{H_3}COOH}} + {n_{{C_2}{H_5}OH}} = \dfrac{1}{2}{n_{{H_2}}} = 0,03 \hfill \\
\Rightarrow {n_{{C_2}{H_5}OH}} = 0,03 - 0,02 = 0,01mol \hfill \\
\Rightarrow m = {m_{{C_2}{H_5}OH}} + {m_{C{H_3}COOH}} = 0,01.46 + 0,02.60 = 1,66g \hfill \\
\end{gathered} $
