$\Delta m_1=5,04g=m_{H_2O}$
$\Rightarrow n_{H_2O}=\dfrac{5,04}{18}=0,28(mol)$
$n_{Cl}=n_{HCl}=n_{AgCl}=\dfrac{5,74}{143,5}=0,04(mol)$
$\Rightarrow n_H=2n_{H_2O}+n_{HCl}=0,6(mol)$
$n_{Ca(OH)_2}=0,2.1,2=0,24(mol)$
$n_{CaCO_3}=0,2(mol)$
- TH1: dư $OH^-$
$\Rightarrow n_C=n_{CaCO_3}=0,2(mol)$
$n_O=\dfrac{5,38-0,6-0,2.12-0,04.35,5}{16}=0,06(mol)$
$n_C : n_H : n_O : n_{Cl}=0,2:0,6:0,06:0,04=10: 30: 3:2$
$\Rightarrow C_{10}H_{30}O_3Cl_2$
- TH2: 2 muối
$n_{Ca(HCO_3)_2}=0,24-0,2=0,04(mol)$
$\Rightarrow n_C=2.0,04+0,2=0,28(mol)$
$m_C+m_H+m_{Cl}=5,38g\Rightarrow $ không có $O$
$n_C: n_H: n_{Cl}=0,28: 0,6:0,04=7:15:1$
$\Rightarrow C_7H_{15}Cl$