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} $