Đáp án:
b) 24,52%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_6}{H_5}OH + 3B{r_2} \to {C_6}{H_2}B{r_3}OH + 3HBr\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
{C_6}{H_5}OH + NaOH \to {C_6}{H_5}ONa + {H_2}O\\
b)\\
n{C_6}{H_2}B{r_3}OH = \dfrac{{16,55}}{{331}} = 0,05\,mol\\
\Rightarrow n{C_6}{H_5}OH = 0,05\,mol\\
nNaOH = 0,15 \times 2 = 0,3\,mol\\
nC{H_3}COOH = 0,3 - 0,05 = 0,25\,mol\\
\% mC{H_3}OH = \dfrac{{52,2 - 0,5 \times 60 - 0,1 \times 94}}{{52,2}} \times 100\% = 24,52\%
\end{array}\)