Giải bài 5 trang 187 SGK Hóa học 11

$\begin{align} & a)\underset{x\,mol}{\mathop{{{C}_{2}}{{H}_{5}}-OH}}\,+Na\to {{C}_{2}}{{H}_{5}}-ONa+\underset{\frac{x}{2}\,mol}{\mathop{\frac{1}{2}{{H}_{2}}}}\, \\ & \underset{y\,mol}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}OH}}\,+Na\to C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-ONa+\underset{\frac{y}{2}\,mol}{\mathop{\frac{1}{2}{{H}_{2}}}}\, \\ & \left\{ \begin{aligned} & 46x+60y=12,2 \\ & \frac{x}{2}+\frac{y}{2}=\frac{2,8}{22,4} \\ \end{aligned} \right. \\ & \Rightarrow \left\{ \begin{aligned} & x=0,2mol \\ & y=0,05mol \\ \end{aligned} \right. \\ \end{align}$

$\%C_2H_5OH=\frac{9,2.100}{12,2}=75,4\%$

$\%C_3H_7OH=24,6\%$

$\begin{align} & b)C{{H}_{3}}-C{{H}_{2}}OH+CuO\xrightarrow{{{t}^{o}}}C{{H}_{3}}-CHO+Cu+{{H}_{2}}O \\ & C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}OH+CuO\xrightarrow{{{t}^{o}}}C{{H}_{3}}-C{{H}_{2}}-CHO+Cu+{{H}_{2}}O \\ \end{align}$