Hỗn hợp E gồm ba este mạch hở

* Hướng dẫn giải

Chọn đáp án D

$+P1:E\stackrel{quydoi}{\to }\underset{\mathrm{12,22}gam}{\underbrace{\left\{\begin{array}{l}{C}_3{H}_{5}COO{C}_3{H}_3:xmol\\ C{H}_{3}OOCCH=CHCOO{C}_3{H}_5:ymol\\ C{H}_2:zmol\end{array}\right\}}}\stackrel{O_2,t^o}{\to }\mathrm{0,37}mol{H}_{2}O$

$+P2:\underset{\mathrm{0,36}mol}{\underbrace{\left\{\begin{array}{l}{C}_3{H}_{5}COO{C}_3{H}_3:kxmol\\ C{H}_{3}OOCCH=CHCOO{C}_3{H}_5:kymol\\ C{H}_2:kzmol\end{array}\right\}}}\stackrel{\mathrm{0,585}molNaOH}{\to }\left\{\begin{array}{l}{C}_3{H}_{3}OH:k\text{x}\\ C_3{H}_{5}OH:ky\\ C{H}_{3}OH:ky\\ C{H}_2:k\text{z}\end{array}\right\}+.$

$\Rightarrow \left\{\begin{array}{l}124\text{x}+170y+14\text{z}=\mathrm{12,22}\\ 4\text{x}+5y+z=\mathrm{0,37}\\ \frac{k\text{x}+2ky}{k\text{x}+ky}=\frac{\mathrm{0,585}}{\mathrm{0,36}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,03}\\ y=\mathrm{0,05}\\ z=0\end{array}\right.\Rightarrow \frac{m_1}{m_2}=\frac{\mathrm{56.0,03}+\mathrm{58.0,05}}{\mathrm{32.0,05}}=\mathrm{2,8625}gannhatvoi\mathrm{2,9.}$