$n_{H_2} = \dfrac{1,68}{22,4} = 0,075(mol)$
$2C_nH_{2n+1}OH + 2Na \to 2C_nH_{2n+1}ONa + H_2$
Theo PTHH :
$n_{ancol} = 2n_{H_2} = 0,15(mol)$
$\Rightarrow M_{ancol} = 14n + 18 = \dfrac{5,92}{0,15} = 39,46$
$\Rightarrow n = 1,53$
Vậy hai ancol là $CH_4O$, $C_2H_6O$
CTCT : $CH_3-O-H$ : Ancol metylic
$CH_3-CH_2-O-H $ : Ancol etylic