Đáp án:

 x=16,56 gam.

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2{C_6}{H_5}OH + 2Na\xrightarrow{{}}2{C_6}{H_5}ONa + {H_2}\)

\(2{C_3}{H_7}OH + 2Na\xrightarrow{{}}2{C_3}{H_7}ONa + {H_2}\)

Ta có: 

\({n_{{H_2}}} = \frac{1}{2}{n_{{C_6}{H_5}OH}} + \frac{1}{2}{n_{{C_3}{H_7}OH}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}}\)

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

\({n_{B{r_2}}} = 0,3.1 = 0,3{\text{ mol}} \to {{\text{n}}_{{C_6}{H_2}B{r_3}OH}} = {n_{{C_6}{H_5}OH}} = \frac{1}{3}{n_{B{r_2}}} = 0,1{\text{ mol}}\)

\( \to m = 0,1.(12.6 + 2 + 80.3 + 17) = 33,1{\text{ gam}}\)

\( \to {n_{{C_3}{H_7}OH}} = 0,3{\text{ mol}} \to {{\text{m}}_{{C_3}{H_7}OH}} = 0,3.74 = 22,2{\text{gam;}}{{\text{m}}_{{C_6}{H_5}OH}} = 0,1.94 = 9,4{\text{ gam}}\)

\( \to \% {m_{{C_3}{H_7}OH}} = \frac{{22,2}}{{22,2 + 9,4}} = 70,25\%  \to \% {m_{{C_6}{H_5}OH}} = 29,75\% \)

Ta có: \({n_{KOH}} = 0,1.1,6 = 0,16{\text{ mol > 0}}{\text{,1}}\) do vậy KOH dư

\({C_6}{H_5}OH + KOH\xrightarrow{{}}{C_6}{H_5}OK + {H_2}O\)

\( \to {n_{{C_6}{H_5}OK}} = 0,1{\text{ mol;}}{{\text{n}}_{KOH{\text{ dư}}}} = 0,16 - 0,1 = 0,06{\text{ mol}}\)

\( \to x = 0,1.(12.6 + 5 + 16 + 39) + 0,06.56 = 16,56{\text{ gam}}\)