Đáp án đúng: D
Giải chi tiết:\(\begin{gathered} 2{C_{\overline n }}{H_{2\overline n + 1}}OH\xrightarrow[{{{140}^0}C}]{{{H_2}S{O_4}\,dac}}{C_{\overline n }}{H_{2\overline n + 1}}O{C_{\overline n }}{H_{2\overline n + 1}} + {H_2}O \hfill \\ 0,15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,075 \hfill \\ \end{gathered} \)
=> Trong a gam ancol : nancol = 4/3. 0,15 = 0,2 (mol)
\(\begin{gathered} 2{C_{\overline n }}{H_{2\overline n + 1}}OH\xrightarrow{{}}\overline n C{O_2} + (\overline n + 1){H_2}O \hfill \\ 0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2\overline n \,\,\,\,\,\,\,0,2.\,(\overline n + 1)\, \hfill \\ {m_{C{O_2}}} - {m_{{H_2}O}} = b + 14,6 - b = 14,6 = 44.\,0,2\overline n \,\,\, - 18.0,2.\,(\overline n + 1) \hfill \\ = > \overline n = 3,5 \hfill \\ \end{gathered} \)
Chọn D