Bảo toàn khối lượng:
$m=m_{\text{ete}}=11-2,7=8,3g$
$n_{H_2O}=\dfrac{2,7}{18}=0,15(mol)$
$\to n_{\text{ancol}}=2n_{H_2O}=0,15.2=0,3(mol)$
$\to \overline{M}_{\text{ancol}}=\dfrac{11}{0,3}=\dfrac{110}{3}$
Đặt CTTQ ancol: $C_nH_{2n+2}O$
$\to 12n+2n+18=\dfrac{110}{3}$
$\to n= 1,33$
Vậy 2 ancol là $CH_3OH$, $C_2H_5OH$