Hai este là đồng phân cấu tạo $C_3H_6O_2$

$n_{hh}=\dfrac{66,6}{74}=0,9(mol)$

Este no, đơn, hở nên ta có:

$n_X=n_{\text{este}}=0,9(mol)$

$2$ ancol $\to$ ete $+H_2O$

$\to n_{H_2O}=\dfrac{n_X}{2}=0,45(mol)$

$\to m=0,45.18=8,1g$

$HCOOC_2H_5+NaOH\xrightarrow{}HCOONa+C_2H_5OH$
$CH_3COOCH_3+NaOH\xrightarrow{}CH_3COONa+CH_3OH$
$2C_2H_5OH\xrightarrow{140^oC,H^+}C_2H_5OC_2H_5+H_2O$
$2CH_3OH\xrightarrow{140^oC,H^+}CH_3OCH_3+H_2O$
$CH_3OH+C_2H_5OH\xrightarrow{140^oC,H^+}CH_3OC_2H_5+H_2O$

$\overline{M}_{este}=74(g/mol)$

( vì cả 2 đều có cùng CTPT là $C_3H_6O_2$) 

$\overline{n}_{este}=\dfrac{66,6}{74}=0,9(mol)=n_{ancol}$
$n_{ancol}=2n_{H_2O}$ ( từ phương trình)

$n_{H_2O}=\dfrac{1}{2}.0,9=0,45(mol)$
$m_{H_2O}=0,45.18=8,1(g)$