Đáp án:

 a) $\begin{gathered}
  \% {m_{{C_6}{H_5}OH}} = 43,12\% ;\% {m_{C{H_3}OH}} = 14,68\%  \hfill \\
  \% {m_{{C_2}{H_5}OH}} = 42,2\%  \hfill \\ 
\end{gathered} $

b) 33,1 g

Giải thích các bước giải:

a) ${n_{{H_2}}} = 0,2mol;{n_{NaOH}} = 0,1mol$

$\begin{gathered}
  2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2} \hfill \\
  2C{H_3}OH + 2Na \to 2C{H_3}ONa + {H_2} \hfill \\
  2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2} \hfill \\
  {C_6}{H_5}OH + NaOH \to {C_6}{H_5}ONa + {H_2}O \hfill \\ 
\end{gathered} $

$ \Rightarrow {n_{{C_6}{H_5}OH}} = {n_{NaOH}} = 0,1mol$

$ \Rightarrow {m_{{C_6}{H_5}OH}} = 0,1.94 = 9,4g$

Gọi x, y là số mol của $C{H_3}OH;{C_2}{H_5}OH$

Ta có hpt: $\left\{ \begin{gathered}
  32x + 46y + 9,4 = 21,8 \hfill \\
  x + y + 0,1 = 0,2.2 \hfill \\ 
\end{gathered}  \right. \Rightarrow \left\{ \begin{gathered}
  x = 0,1 \hfill \\
  y = 0,2 \hfill \\ 
\end{gathered}  \right.$

$\begin{gathered}
   \Rightarrow \% {m_{{C_6}{H_5}OH}} = \dfrac{{9,4}}{{21,8}}.100\%  = 43,12\%  \hfill \\
  \% {m_{C{H_3}OH}} = \dfrac{{32.0,1}}{{21,8}}.100\%  = 14,68\%  \hfill \\
  \% {m_{{C_2}{H_5}OH}} = 100 - 14,68 - 43,12 = 42,2\%  \hfill \\ 
\end{gathered} $

b) ${C_6}{H_5}OH + 3B{r_2} \to {C_6}{H_3}B{r_3}O \downarrow  + 3HBr$

$\begin{gathered}
  {n_{{C_6}{H_5}OH}} = {n_{{C_6}{H_3}B{r_3}O}} = 0,1mol \hfill \\
   \Rightarrow {m_{{C_6}{H_3}B{r_3}O}} = 0,1.331 = 33,1g \hfill \\ 
\end{gathered} $

c) Những ete có thể thu được là:

$C{H_3} - O - C{H_3};{C_2}{H_5} - O - {C_2}{H_5};C{H_3} - O - {C_2}{H_5}$