a,
$\Delta m_1=7,2g= m_{H_2O}$
$\Rightarrow n_H=2n_{H_2O}=\dfrac{2.7,2}{18}=0,8(mol)$
$Ca(HCO_3)_2\buildrel{{t^o}}\over\to CaCO_3+CO_2+H_2O$
$n_{CaCO_3(1)}=\dfrac{10}{100}=0,1(mol)$
$n_{CaCO_3(2)}=\dfrac{10}{100}=0,1(mol)=n_{Ca(HCO_3)_2}$
$CO_2+Ca(OH)_2\to CaCO_3+H_2O$
$2CO_2+Ca(OH)_2\to Ca(HCO_3)_2$
$\Rightarrow n_{CO_2}=0,1+0,1.2=0,3(mol)$
$\Rightarrow n_C=0,3(mol)$
BTKL, suy ra $n_O=\dfrac{6-0,3.12-0,8}{16}=0,1(mol)$
b,
$\%m_C=\dfrac{0,3.12.100}{6}=60\%$
$\%m_H=\dfrac{0,8.100}{6}=13,33\%$
$\%m_O=26,67\%$
c,
$n_C : n_H : n_O=0,3:0,8:0,1=3:8:1$
d,
CTĐGN: $(C_3H_8O)_n$
$M=60\Rightarrow n=1$
$\Rightarrow$ CTPT $C_3H_8O$