Đáp án:

 3,31 g

Giải thích các bước giải:

\(\begin{array}{l}
{C_6}{H_5}OH + KOH \to {C_6}{H_5}OK + {H_2}O\\
C{H_3}COOH + KOH \to C{H_3}COOK + {H_2}O\\
2C{H_3}COOH + N{a_2}C{O_3} \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
nC{O_2} = \dfrac{{0,224}}{{22,4}} = 0,01\,mol\\
 \Rightarrow nC{H_3}COOH = 2nC{O_2} = 0,02\,mol\\
nKOH = 0,3 \times 0,1 = 0,03\,mol\\
n{C_6}{H_5}OH = 0,03 - 0,02 = 0,01\,mol\\
\% m{C_6}{H_5}OH = \dfrac{{0,01 \times 94}}{{3,52}} \times 100\%  = 26,7\% \\
\% mC{H_3}COOH = \dfrac{{0,02 \times 60}}{{3,52}} \times 100\%  = 34,1\% \\
\% m{C_2}{H_5}OH = 100 - 34,1 - 26,7 = 39,2\% \\
b)\\
{C_6}{H_5}OH + 3B{r_2} \to {C_6}{H_2}B{r_3}OH + 3HBr\\
n{C_6}{H_2}B{r_3}OH = n{C_6}{H_5}OH = 0,01\,mol\\
m{C_6}{H_2}B{r_3}OH = 0,01 \times 331 = 3,31g
\end{array}\)