Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{C{H_3}OH}} = 40,51\% \\
\% {m_{{C_6}{H_5}OH}} = 59,49\% \\
b)\\
{m_{{C_6}{H_2}B{r_3}OH}} = 33,1g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2C{H_3}OH + 2Na \to 2C{H_3}ONa + {H_2}\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
hh:C{H_3}OH(a\,mol),{C_6}{H_5}OH(b\,mol)\\
\left\{ \begin{array}{l}
32a + 94b = 15,8\\
0,5a + 0,5b = 0,15
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,1\\
\% {m_{C{H_3}OH}} = \dfrac{{0,2 \times 32}}{{15,8}} \times 100\% = 40,51\% \\
\% {m_{{C_6}{H_5}OH}} = 100 - 40,51 = 59,49\% \\
b)\\
{C_6}{H_5}OH + 3B{r_2} \to {C_6}{H_2}B{r_3}OH + 3HBr\\
{n_{{C_6}{H_2}B{r_3}OH}} = {n_{{C_6}{H_5}OH}} = 0,1\,mol\\
{m_{{C_6}{H_2}B{r_3}OH}} = 0,1 \times 331 = 33,1g
\end{array}\)
