Đáp án:
a) 39,22 và 60,78%
b) 39,72g
Giải thích các bước giải:
\(\begin{array}{l}
2)\\
a)\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2}\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
n{H_2} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
nhh = 0,25 \times 2 = 0,5\,mol\\
hh:{C_6}{H_5}OH(a\,mol),{C_2}{H_5}OH(b\,mol)\\
94a + 46b = 28,76\\
a + b = 0,5\\
\Rightarrow a = 0,12;b = 0,38\\
\% m{C_6}{H_5}OH = \dfrac{{0,12 \times 94}}{{28,76}} \times 100\% = 39,22\% \\
\% m{C_2}{H_5}OH = 100 - 39,22 = 60,78\% \\
b)\\
{C_6}{H_5}OH + 3B{r_2} \to {C_6}{H_2}B{r_3}OH + 3HBr\\
n{C_6}{H_2}B{r_3}OH = n{C_6}{H_5}OH = 0,12\,mol\\
m{C_6}{H_2}B{r_3}OH = 0,12 \times 331 = 39,72g
\end{array}\)
