Đáp án:
\(\begin{array}{l}
a)\\
{m_{{C_2}{H_5}OH}} = 11,04g\\
{m_{{C_6}{H_5}OH}} = 5,64g\\
b)\\
\% {m_{{C_2}{H_5}OH}} = 66,19\% \\
\% {m_{{C_6}{H_5}OH}} = 33,81\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}(1)\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2}(2)\\
{C_6}{H_5}OH + 3B{r_2} \to {C_6}{H_2}B{r_3}OH + 3HBr\\
{n_{{C_6}{H_2}B{r_3}OH}} = \dfrac{{19,86}}{{331}} = 0,06\,mol\\
{n_{{C_6}{H_5}OH}} = {n_{{C_6}{H_2}B{r_3}OH}} = 0,06\,mol\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{{H_2}(2)}} = \dfrac{{0,06}}{2} = 0,03\,mol\\
{n_{{H_2}(1)}} = 0,15 - 0,03 = 0,12\,mol\\
{n_{{C_2}{H_5}OH}} = 0,12 \times 2 = 0,24\,mol\\
{m_{{C_2}{H_5}OH}} = 0,24 \times 46 = 11,04g\\
{m_{{C_6}{H_5}OH}} = 0,06 \times 94 = 5,64g\\
b)\\
\% {m_{{C_2}{H_5}OH}} = \dfrac{{11,04}}{{11,04 + 5,64}} \times 100\% = 66,19\% \\
\% {m_{{C_6}{H_5}OH}} = 100 - 66,19 = 33,81\%
\end{array}\)
