Đáp án:

 a) m = 33,2 g

$\% {m_{{C_2}{H_5}OH}} = 27,71\% ;\% {m_{C{H_3}COOH}} = 72,29\% $

b) 10,912 g

Giải thích các bước giải:

 a) - Trong mỗi phần:

${n_{C{H_3}COOH}} = {n_{NaOH}} = 0,2mol$

$\begin{gathered}
  {n_{C{H_3}COOH}} + {n_{{C_2}{H_5}OH}} = 2{n_{{H_2}}} = 2.0,15 = 0,3mol \hfill \\
   \Rightarrow {n_{{C_2}{H_5}OH}} = 0,3 - 0,2 = 0,1mol \hfill \\ 
\end{gathered} $

$\begin{gathered}
   \Rightarrow {m_A} = {m_{{C_2}{H_5}OH}} + {m_{C{H_3}COOH}} \hfill \\
   = 2.\left( {0,1.46 + 0,2.60} \right) = 33,2g \hfill \\ 
\end{gathered} $

$\begin{gathered}
   \Rightarrow \% {m_{{C_2}{H_5}OH}} = \frac{{0,2.46}}{{33,2}}.100\%  = 27,71\%  \hfill \\
   \Rightarrow \% {m_{C{H_3}COOH}} = 100 - 27,71 = 72,29\%  \hfill \\ 
\end{gathered} $

b) 

$C{H_3}COOH + {C_2}{H_5}OH\overset {{H_2}S{O_{4d}},{t^o}} \leftrightarrows C{H_3}COO{C_2}{H_5} + {H_2}O$

Do ${n_{C{H_3}COOH}} > {n_{{C_2}{H_5}OH}}$ ⇒ Tính theo $_{{C_2}{H_5}OH}$

$\begin{gathered}
  H = 62\%  \Rightarrow {n_{{C_2}{H_5}OH(pu)}} = 0,2.H = 0,124mol \hfill \\
   \Rightarrow {n_{este}} = {n_{{C_2}{H_5}OH(pu)}} = 0,124mol \hfill \\
   \Rightarrow {m_{este}} = 0,124.88 = 10,912g \hfill \\ 
\end{gathered} $