$n_{K_2CO_3}=0.46(mol)$

Bảo toàn nguyên tố $K:$ `=>`$n_{COOH}=n_{KOH}=0.92(mol)$

`=>`$n_{H_2O/Trung \,\,  hòa}=0.92(mol)$

Bảo toàn khối lượng:

$m_A+m_{KOH}=m_{D}+m_{H_2O/Trung \,\,  hòa}$

`=>`$m_D=81(gam)$

Bảo toàn khối lượng:

$m_D+m_{O_2}=m_{K_2CO_3}+m_{CO_2}+m_{H_2O}$

`=>`$n_{O_2}=0.83(mol)$

$n_{CO_2}=a(mol);n_{H_2O}=b(mol)$

`->`$44a+18b=44.08$ $(1)$

Bảo toàn nguyên tố $O:$

$2n_{COOH}+2n_{O_2}=2n_{CO_2}+n_{H_2O}+3n_{K_2CO_3}$

`<=>`$2a+b=2.12$ $(2)$

Từ $(1),(2)$ suy ra: $a=0.74;b=0.64$

`=>`$n_X=a-b=0.1(mol)$  (hệ quả ở dưới cuối phần trả lời)

`=>`$n_{Y+Z}=n_{COOH}-n_{X}=0.82(mol)$

Bảo toàn  nguyên tố $C:$

$n_C=n_{K_2CO_3}+n_{CO_2}=1.2(mol)$

$\Delta n_C=n_C-(3n_X+n_{Y+Z})=0.08(mol)$

Do đó,

$X:C_2H_3COOH:0.1(mol)$ `=>`$\%m_A=15.64\%$

$Y:HCOOH:0.82-0.08=0.74(mol)$

$Z:CH_3COOH:0.08(mol)$

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$\rm C_nH_{2n+1}COOK+\dfrac{6n+2}{4}O_2 \xrightarrow{t^o}\dfrac{1}{2}K_2CO_3+(n+0.5)CO_2+(n+0.5)H_2O$

$\rm C_nH_{2n-1}COOK+\dfrac{6n}{4}O_2 \xrightarrow{t^o}\dfrac{1}{2}K_2CO_3+(n+0.5)CO_2+(n-0.5)H_2O$

`=>`$n_{CO_2}-n_{H_2O}=n+0.5-n-(-0.5)=1(mol)=n_{acid ,1C=C}=n_X$