Đáp án:

\( \% {m_{{C_6}{H_5}OH}} = 57,82\% ; \% {m_{{C_2}{H_5}OH}} = 42,18\% \)

\( {m_{{C_6}{H_2}B{r_3}OH}}  = 46,34{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\({C_6}{H_5}OH + NaOH\xrightarrow{{}}{C_6}{H_5}ONa + {H_2}O\) 

Ta có:

\({n_{NaOH}} = 0,35.0,4 = 0,14{\text{ mol = }}{{\text{n}}_{{C_6}{H_5}OH}}\)

\( \to {m_{{C_6}{H_5}OH}} = 0,14.94 = 13,16{\text{ gam}}\)

\( \to \% {m_{{C_6}{H_5}OH}} = \frac{{13,16}}{{22,76}} = 57,82\%  \to \% {m_{{C_2}{H_5}OH}} = 42,18\% \)

Cho hỗn hợp tác dụng với \(Br_2\)

\({C_6}{H_5}OH + 3B{r_2}\xrightarrow{{}}{C_6}{H_2}B{r_3}OH + 3HBr\)

Ta có:

\({n_{{C_6}{H_2}B{r_3}OH}} = {n_{{C_6}{H_5}OH}} = 0,14{\text{ mol}}\)

\( \to {m_{{C_6}{H_2}B{r_3}OH}} = 0,14.(12.6 + 2 + 80.3 + 17) = 46,34{\text{ gam}}\)