$n_E=\dfrac{16,32}{136}=0,12 mol$
$\dfrac{n_{NaOH}}{n_E}=1,67\to$ có 1 este phenol
Tạo 2 muối nên este là $HCOOC_6H_4CH_3$ (x mol), $HCOOCH_2C_6H_5$ (y mol)
$\Rightarrow x+y=0,12; 2x+y=0,2$
$\Rightarrow x=0,08; y=0,04$
Muối gồm:
$HCOONa: x+y=0,12$
$CH_3C_6H_4ONa: 0,08 mol$
$\to m=0,12.68+0,08.130=18,56g$