$\Delta m_1=10,8g=m_{H_2O}$
$\Rightarrow n_{H_2O}=\dfrac{10,8}{18}=0,6(mol)$
$n_{CaCO_3(1)}=\dfrac{18}{100}=0,18(mol)$
$n_{CaCO_3(2)}=\dfrac{30}{100}=0,3(mol)$
$Ca(HCO_3)_2+Ca(OH)_2\to 2CaCO_3+2H_2O$
$\Rightarrow n_{Ca(HCO_3)_2}=0,15(mol)$
$CO_2+Ca(OH)_2\to CaCO_3+H_2O$
$2CO_2+Ca(OH)_2\to Ca(HCO_3)_2$
$\Rightarrow n_{CO_2}=n_{CaCO_3}+2n_{Ca(HCO_3)_2}=0,48(mol)$
$\Rightarrow n_A=n_{H_2O}-n_{CO_2}=0,12(mol)$
Số C trong A: $\dfrac{n_{CO_2}}{n_A}=4$
Vậy CTPT A là $C_4H_{10}$
CTCT:
(1) $CH_3-CH_2-CH_2-CH_3$ (butan)
(2) $CH_3-CH(CH_3)-CH_3$ (isobutan)