$m_{H_2O}=3,6g$
$\Rightarrow n_H=2n_{H_2O}=\dfrac{2.3,6}{18}=0,4(mol)$
$n_{CaCO_3(1)}=\dfrac{10}{100}=0,1(mol)$
$n_{CaCO_3(2)}=\dfrac{20}{100}=0,2(mol)$
$Ca(HCO_3)_2+Ca(OH)_2\to 2CaCO_3+2H_2O$
$\Rightarrow n_{Ca(HCO_3)_2}=0,1(mol)$
Bảo toàn C:
$n_C=n_{CO_2}=2n_{Ca(HCO_3)_2}+n_{CaCO_3(1)}$
$=0,1.2+0,1=0,3(mol)$
$\Rightarrow n_O=\dfrac{7,2-0,3.12-0,4}{16}=0,2(mol)$
$n_C: n_H : n_O=0,3:0,4:0,2=3:4:2$
$\Rightarrow$ CTĐGN $(C_3H_4O_2)_n$
$M=36.2=72\Rightarrow n=1$
Vậy CTPT là $C_3H_4O_2$
