Đáp án:

Bảo toàn nguyên tố C:

$\rm n_{CO_2}=n_{CaCO_3}=\dfrac{76}{100}=0,76\ (mol)$

$\rm n_{H_2O}=\dfrac{-30,08+76-0,76×44}{18}=0,66\ (mol)$

Bảo toàn nguyên tố H:

$\rm n_H=2n_{H_2O}=1,32\ (mol)$

$\rm n_O=\dfrac{20,04-0,76×44-1,32}{18}=0,6\ (mol)$

$\rm n_{COONa}=n_{COO}=\dfrac{1}{2}n_{O}=0,3\ (mol)$

$\rm M_{muoi}=\dfrac{22,08}{0,03}=73,6\ (g/mol)$

Vậy 2 muối là: $\begin{cases}\rm HCOONa:a\ (mol)\\\rm CH_3COONa:b\ (mol)\end{cases}$

Hệ phương trình:

$\begin{cases}\rm a+b=0,3\\\rm 68a+82b=22,08\end{cases}\to\begin{cases}\rm a=0,18\\\rm b=0,12\end{cases}$

Bảo toàn khối lượng:

$\rm m_{ancol}=20,04+0,3×40-22,08=9,96\ (g)$

Đồng đẳng hoá ancol: $\begin{cases}\rm CH_3OH:0,1\ (mol)\\\rm C_2H_4(OH)_2:0,1\ (mol)\\\rm CH_2:0,04\end{cases}$

$\rm n_{CH_2}=\dfrac{9,96-0,1×(32+62)}{14}=0,04\ (mol)$

Ghép ancol: $\begin{cases}\rm CH_3OH:0,08\\\rm C_3H_7OH:0,02\\\rm C_2H_4(OH)_2:0,1\end{cases}$

Ghép este: $\begin{cases}\rm HCOO-CH_3COOC_2H_4:0,1\ (mol)\\\rm HCOOCH_3:0,08\ (mol)\\\rm CH_3COOC_3H_7:0,02\ (mol)\end{cases}$

Khối lượng mỗi chất:

$\rm m_{HCOO-CH_3COOC_2H_4}=0,1×132=13,2\ (g)\\m_{HCOOCH_3}=0,08×60=4,8\ (g)\\m_{CH_3COOC_3H_7}=0,02×102=2,04\ (g)$