$n_C=n_{CO_2}=n_{CaCO_3}=\dfrac{48}{100}=0,48(mol)$
$\Delta m_b=m_{CO_2}+m_{H_2O}$
$\Rightarrow m_{H_2O}=27,6-0,48.44=6,48g$
$\Rightarrow n_H=2n_{H_2O}=0,36.2=0,72(mol)$
$m_C+m_H=0,48.12+0,72=6,48g$
$\Rightarrow n_O=\dfrac{10,32-6,48}{16}=0,24(mol)$
$n_C : n_H : n_O=0,48: 0,72: 0,24=2:3:1$
$\Rightarrow$ CTĐGN $(C_2H_3O)_n$
$M-A=43.2=86\Rightarrow n=2$
Vậy A là $C_4H_6O_2$