Question

Đã trả lời bởi chuyên gia

Answer 1

Đáp án D

Ta thấy chất béo tạo từ 2 gốc axit $\{\begin{cases} C_{17} H_{33} C O O H \\ C_{17} H_{31} C O O H \end{cases} \overset{C T P T c h u n g}{\to} (C_{17} H_{y} C O O)_{3} C_{3} H_{5} \underset{n_{g l y x e r o l}}{\underset{⏟}{a m o l}}$

BTNT C $\underset{18.3 + 3}{\underset{⏟}{57}} n_{X} = n_{C O_{2}} \Rightarrow n_{C O_{2}} = 57 a m o l$

Dùng CT liên hệ $n_{C O_{2}} + n_{N_{2}} - n_{H_{2} O} = (k - 1) . n_{X} \Rightarrow \underset{57 a}{\underset{⏟}{n_{C O_{2}}}} - \underset{20, 15}{\underset{⏟}{n_{H_{2} O}}} = (k - 1) . \underset{a}{\underset{⏟}{n_{X}}} \Leftrightarrow (58 - k) a = 20, 15 (1)$

chất béo X có $<\begin{array}{l} 3 l k π t r o n g - C O O - \\ k - 3 l k π t r o n g - C - C - \end{array}$

Ta có: $\underset{k - 3}{\underset{⏟}{l k π t r o n g - C - C -}} = \underset{a}{\underset{⏟}{\frac{{\overset{⏞}{n_{B r_{2}}}}^{4, 625 a}}{n_{X}}}} \Rightarrow k - 3 = 4, 625 \Rightarrow k = 7, 625 \overset{t h a y v ̀ a o (1)}{\to} a = 0, 4 m o l$

BTKL trong X: $m_{X} = \underset{\overset{B T N T C}{\to} 57.0, 4.12}{\underset{⏟}{m_{C}}} + \underset{\overset{B T N T H}{\to} 20, 15.2}{\underset{⏟}{m_{H}}} + \underset{0, 4.6.16}{\underset{⏟}{m_{O}}} \Rightarrow m_{X} = 352, 3 g a m$

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Answer 2