Hỗn hợp E gồm X là este no, hai chức; Y là este tạo bởi glixerol và một axit cacboxylic đơn chức, không no chứa một liên kết C=C

* Hướng dẫn giải

$\begin{array}{l}+\left\{\begin{array}{l}X:esteno,haichức\\ Y:C_3{H}_5{\left(OOC{C}_n{H}_{2n-1}\right)}_3\end{array}\right.\stackrel{NaOH}{\to }\left\{\begin{array}{l}3muối\\ 2ancolcùngsốC\end{array}\right.\Rightarrow 2ancollà\left\{\begin{array}{l}{C}_3{H}_5{\left(OH\right)}_3\\ C_3{H}_6{\left(OH\right)}_2\end{array}\right.\\ +\text{\hspace{0.17em}\hspace{0.17em}Ứ}ngvới0,09molE:\left\{\begin{array}{l}{n}_X+n_Y=0,09\\ n_{NaOH}=2n_X+3n_Y=0,24\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_X=0,03\\ n_Y=0,06\end{array}\right.\Rightarrow \frac{n_X}{n_Y}=\frac{1}{2}\end{array}$

$\begin{array}{l}+14,76gamE\stackrel{quy\text{\hspace{0.17em}\hspace{0.17em}đổ}i}{\to }\left\{\begin{array}{l}{C}_3{H}_6{\left(OOCC{H}_3\right)}_2:xmol\\ C_3{H}_5{(OOCCH=C{H}_2)}_3:2xmol\\ C{H}_2:ymol\end{array}\right\}\Rightarrow \left\{\begin{array}{l}160x+2x.254+14y=14,76\\ n_{C{O}_2}=7x+2x.12+y=0,72\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,02\\ y=0,1\end{array}\right.\\ \Rightarrow y=2x.3-x\Rightarrow Egồm\left\{\begin{array}{l}XlàC_3{H}_6{\left(OOCC{H}_3\right)}_2-\left(C{H}_2\right)\iff HCOO{C}_3{H}_{6}OOCC{H}_3:0,02mol\\ YlàC_3{H}_5{(OOCCH=C{H}_2)}_3+\left(3C{H}_2\right)\iff C_3{H}_5{\left(OOC{C}_3{H}_5\right)}_3:0,04mol\end{array}\right\}\\ \Rightarrow \%Y=\frac{0,04.296}{14,76}=\boxed{80,22\%}\end{array}$

Chọn A