Giải bài 6 trang 195 SGK Hóa học 11

$\begin{align} & \underset{x\,mol}{\mathop{a){{C}_{2}}{{H}_{5}}OH}}\,+Na\to {{C}_{2}}{{H}_{5}}ONa+\underset{\frac{x}{2}mol}{\mathop{\frac{1}{2}{{H}_{2}}}}\, \\ & \underset{y\,mol}{\mathop{{{C}_{6}}{{H}_{5}}OH}}\,+Na\to {{C}_{6}}{{H}_{5}}ONa+\underset{\frac{y}{2}mol}{\mathop{\frac{1}{2}{{H}_{2}}}}\, \\ & \underset{94g}{\mathop{{{C}_{6}}{{H}_{5}}OH}}\,+3B{{r}_{2}}\to \underset{331g}{\mathop{{{C}_{6}}{{H}_{2}}(OH)B{{r}_{3}}}}\,+3HBr \\ & b){{n}_{phenol}}={{n}_{\downarrow }}=\frac{19,86}{331}=0,06mol \\ & {{m}_{phenol}}=0,06.94=5,04g \\ & \frac{x}{2}+\frac{y}{2}=\frac{3,36}{22,4} \\ & \Leftrightarrow x+y=0,3 \\ \end{align}$

$\Rightarrow \left\{ \begin{align} & y=0,06 \\ & x=0,3-0,06=0,24 \\ \end{align} \right.$

$\begin{align} & {{m}_{etan ol}}=0,24.46=11,04g \\ & {{m}_{hh}}=11,04+5,64=16,68g \\ \end{align}$

$\%etanol=\frac{11,04.100}{16,68}=66,2\%$

$\%phenol=33,8\%$