Đốt cháy hoàn toàn x mol este E chỉ chứa chức este cần dùng 3,5x mol O2,

Đáp án đúng là: A

$$\begin{gathered} {\mathrm{n}}_{\mathrm{\pi }}={\mathrm{n}}_{{\mathrm{CO}}_2}-{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}+{\mathrm{n}}_{\mathrm{E}}=2\mathrm{x} \mathrm{mol} \\ \to \mathrm{E} \mathrm{chứa} 2 \mathrm{liên} \mathrm{kết} \mathrm{\pi } \to \mathrm{E} \mathrm{đơn} \mathrm{chức}, \mathrm{có} 1 \mathrm{nối} \mathrm{đôi} \mathrm{C} = \mathrm{C} \mathrm{hoặc} \mathrm{E} \mathrm{là} \mathrm{este} \mathrm{no}, 2 \mathrm{chức}. \\ \to \left[\begin{array}{l}{\mathrm{TH}}_1:\mathrm{CT}\mathrm{E}:{\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}-2}{\mathrm{O}}_2\\ {\mathrm{TH}}_2:\mathrm{CT}\mathrm{E}:{\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}-2}{\mathrm{O}}_4\end{array}\right. \\ \mathrm{TH}1: {\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}-2}{\mathrm{O}}_2+\frac{{\displaystyle 3\mathrm{n}-3}}{{\displaystyle 2}}{\mathrm{O}}_2\stackrel{{\mathrm{t}}^{\mathrm{o}}}{\to }{\mathrm{nCO}}_2+(\mathrm{n}-1){\mathrm{H}}_2\mathrm{O} \\ \Rightarrow \frac{{\displaystyle 3\mathrm{n}-3}}{{\displaystyle 2}}=\frac{{\displaystyle {\mathrm{n}}_{{\mathrm{O}}_2}}}{{\displaystyle {\mathrm{n}}_{\mathrm{E}}}}=3,5\Rightarrow \mathrm{n}=3,3 \left(\mathrm{loại}\right) \\ \mathrm{TH}2: {\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}-2}{\mathrm{O}}_4+\frac{{\displaystyle 3\mathrm{n}-5}}{{\displaystyle 2}}{\mathrm{O}}_2\stackrel{{\mathrm{t}}^{\mathrm{o}}}{\to }{\mathrm{nCO}}_2+(\mathrm{n}-1){\mathrm{H}}_2\mathrm{O} \\ \Rightarrow \frac{{\displaystyle 3\mathrm{n}-5}}{{\displaystyle 2}}=\frac{{\displaystyle {\mathrm{n}}_{{\mathrm{O}}_2}}}{{\displaystyle {\mathrm{n}}_{\mathrm{E}}}}=3,5\Rightarrow \mathrm{n}=4 \\ \to \mathrm{CT}: {\left({\mathrm{COOCH}}_3\right)}_2 ; {\left(\mathrm{HCOO}\right)}_2{\mathrm{C}}_2{\mathrm{H}}_4; \mathrm{HCOOCH}\left({\mathrm{CH}}_3\right)\mathrm{OOCH}; {\mathrm{HCOOCH}}_2{\mathrm{COOCH}}_3 \end{gathered}$$