Thuỷ phân hoàn toàn m gam hỗn hợp X gồm 2 triglixerit thu được hỗn

* Hướng dẫn giải

Đáp án D

Ta thấy chất béo tạo từ 2 gốc axit $\left\{\begin{array}{l}{C}_{17}{H}_{33}COOH\\ C_{17}{H}_{31}COOH\end{array}\right.\stackrel{CTPT chung}{\to }(C_{17}{H}_{y}COO{)}_3{C}_3{H}_5 \underset{n_{glyxerol}}{\underbrace{a mol}}$

BTNT C $\underset{18.3+3}{\underbrace{57}}{n}_X=n_{C{O}_2}\Rightarrow n_{C{O}_2}=57a mol$

Dùng CT liên hệ $n_{C{O}_2}+n_{N_2}-n_{H_{2}O}=(k-1).n_X \Rightarrow \underset{57a}{\underbrace{n_{C{O}_2}}}-\underset{20,15}{\underbrace{n_{H_{2}O}}}=(k-1).\underset{a}{\underbrace{n_X}}\iff (58-k)a=20,15 \left(1\right)$

chất béo X có 

Ta có: $\underset{k-3}{\underbrace{lk \pi trong -C-C-}}=\underset{a}{\underbrace{\frac{{\overbrace{n_{B{r}_2}}}^{4,625a}}{n_X}}}\Rightarrow k-3=4,625\Rightarrow k=7,625 \stackrel{thay v̀ao \left(1\right)}{\to }a=0,4 mol$

BTKL trong X: $m_X=\underset{\stackrel{BTNT C}{\to }57.0,4.12}{\underbrace{m_C}}+\underset{\stackrel{BTNT H}{\to }20,15.2}{\underbrace{m_H}}+\underset{0,4.6.16}{\underbrace{m_O}}\Rightarrow m_X=352,3 gam$