Đáp án:

Gọi cacboxylic acid là $\rm RCOOH$.

Phản ứng xảy ra:

$\rm 2RCOOH+CaCO_3\to (RCOO)_2Ca+CO_2+H_2O$

Theo phương trình:

$\rm\kern,9cm n_{RCOOH}=2n_{(RCOO)_2Ca}\\\Leftrightarrow\kern,2cm\dfrac{2,88}{M_R+45}=\dfrac{2×3,64}{2M_R+128}\\\Leftrightarrow\kern,2cmM_R=27\ (g/mol)\ (CH_2=CH-)$

Vậy X có CTCT là: $\rm CH_2=CH-COOH$

X: RCOOH 

`2RCOOH + CaCO_3→ (RCOO)_2Ca + CO_2+ H_2O`

 Theo pt: 

`n_{RCOOH}= 2n_{(RCOO)_2Ca}`

`<=> {2,88}/{R + 45}=2 {3,64}/{2R + 128}`

`→ R=27`$(g/mol)$

`→ R: C_2H_3-`

`→ X: C_2H_3COOH`

CTCT: `CH_2=CH-COOH`