Đáp án:
C2H3COOH: 58%; CH3COOH: 12,16%; C2H5COOH: 29,84%
Giải thích các bước giải:
a)
$\begin{gathered}
C{H_2} = CH - COOH + B{r_2} \to C{H_2}Br - CHBr - COOH \hfill \\
C{H_2} = CH - COOH + NaOH \to C{H_2} = CH - COONa + {H_2}O \hfill \\
{C_2}{H_5}COOH + NaOH \to {C_2}{H_5}COONa + {H_2}O \hfill \\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O \hfill \\
\end{gathered} $
b)
$\begin{gathered}
{n_{B{r_2}}} = \dfrac{{6,4}}{{160}} = 0,04mol \Rightarrow {n_{{C_2}{H_3}COOH}} = 0,04mol \hfill \\
\Rightarrow {m_{{C_2}{H_3}COOH}} = 0,04.72 = 2,88g \hfill \\
\Rightarrow \% {m_{{C_2}{H_3}COOH}} = \dfrac{{2,88}}{{4,96}}.100\% = 58\% \hfill \\
\end{gathered} $
Gọi x, y là số mol của ${C_2}{H_5}COOH;C{H_3}COOH$
ta có hpt: $\left\{ \begin{gathered}
x + y + 0,04 = 0,07 \hfill \\
74x + 60y + 2,88 = 4,96 \hfill \\
\end{gathered} \right. \Rightarrow \left\{ \begin{gathered}
x = 0,02 \hfill \\
y = 0,01 \hfill \\
\end{gathered} \right.$
$\begin{gathered}
\Rightarrow \% {m_{{C_2}{H_5}COOH}} = \dfrac{{0,02.74}}{{4,96}}.100\% = 29,84\% \hfill \\
\Rightarrow \% {m_{C{H_3}COOH}} = 100 - 29,84 - 58 = 12,16\% \hfill \\
\end{gathered} $
