Hỗn hợp X gồm metyl fomat, đimetyl oxalat, glixerol triaxetat và phenyl axetat.

Đáp án đúng là: D

X: HCOOCH₃; (COOCH₃)₂; (CH₃COO)₃C₃H­₅; CH₃COOC₆H_5­.

Quy đổi hỗn hợp về: $\left\{\begin{array}{l}\mathrm{COO}:\mathrm{x}\\ {\mathrm{CH}}_2:\mathrm{y}\\ {\mathrm{H}}_2:\mathrm{z}={\mathrm{n}}_{\mathrm{hh}}-4.{\mathrm{n}}_{{\mathrm{CH}}_3{\mathrm{COOC}}_6{\mathrm{H}}_5}\end{array}\right.$

-> 44x + 14y + 2z = 47,3

$\left\{\begin{array}{l}\mathrm{COO}:\mathrm{x}\\ {\mathrm{CH}}_2:\mathrm{y}\\ {\mathrm{H}}_2:\mathrm{z}\end{array}\right.+{\mathrm{O}}_2\stackrel{{\mathrm{t}}^{\mathrm{o}}}{\to }\left\{\begin{array}{l}{\mathrm{CO}}_2:2,1\\ {\mathrm{H}}_2\mathrm{O}:1,45\end{array}\right.$

-> BT C: x + y = 2,1

-> BT H: y + z = 1,45

->  x = 0,6; y = 1,5; z = -0,05

 $\to {\mathrm{n}}_{\mathrm{COO}}={\mathrm{n}}_{\mathrm{OH}\left(\mathrm{ancol}\right)}+{\mathrm{n}}_{{\mathrm{CH}}_3{\mathrm{COOC}}_6{\mathrm{H}}_5}=2.{\mathrm{n}}_{{\mathrm{H}}_2}+{\mathrm{n}}_{{\mathrm{CH}}_3{\mathrm{COOC}}_6{\mathrm{H}}_5}$

-> 0,6 = 0,5 + ${\mathrm{n}}_{{\mathrm{CH}}_3{\mathrm{COOC}}_6{\mathrm{H}}_5} \to {\mathrm{n}}_{{\mathrm{CH}}_3{\mathrm{COOC}}_6{\mathrm{H}}_5}$  = 0,1

->  ${\mathrm{n}}_{\mathrm{NaOH}}={\mathrm{n}}_{\mathrm{COO}}+{\mathrm{n}}_{{\mathrm{CH}}_3{\mathrm{COOC}}_6{\mathrm{H}}_5}=0,7;{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}={\mathrm{n}}_{{\mathrm{CH}}_3{\mathrm{COOC}}_6{\mathrm{H}}_5}=0,1$

BTKL: 47,3 + 0,7.40 = 15,6 + m + 0,1.18

-> m = 57,9g