Đáp án:

a) 32,86% và 67,14%

 b) 33,1g

c) 100ml

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2}\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
n{H_2} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
 \Rightarrow nhh = 0,1 \times 2 = 0,2\,mol\\
hh:{C_2}{H_5}OH(a\,mol),{C_6}{H_5}OH(b\,mol)\\
46a + 94b = 14\\
a + b = 0,2\\
 \Rightarrow a = b = 0,1\\
\% m{C_2}{H_5}OH = \dfrac{{0,1 \times 46}}{{14}} \times 100\%  = 32,86\% \\
\% m{C_6}{H_5}OH = 100 - 32,86 = 67,14\% \\
b)\\
{C_6}{H_5}OH + 3B{r_2} \to {C_6}{H_2}B{r_3}OH + 3HBr\\
n{C_6}{H_2}B{r_3}OH = n{C_6}{H_5}OH = 0,1\,mol\\
m{C_6}{H_2}B{r_3}OH = 0,1 \times 331 = 33,1g\\
c)\\
{C_6}{H_5}OH + NaOH \to {C_6}{H_5}ONa + {H_2}O\\
nNaOH = n{C_6}{H_5}OH = 0,1\,mol\\
VNaOH = \dfrac{{0,1}}{1} = 0,1l = 100ml
\end{array}\)