Câu 1:
a,
Đặt CTTQ hai ancol là $C_nH_{2n+2}O$
$n_{CO_2}=\dfrac{2,016}{22,4}=0,09(mol)$
$\to n_{\rm ancol}=\dfrac{n_{CO_2}}{n}=\dfrac{0,09}{n}(mol)$
$\to \overline{M}_{\rm ancol}=\dfrac{1,89n}{0,09}=21n$
$\to 14n+18=21n$
$\to n=\dfrac{18}{7}\approx 2,6$
Vậy CTPT hai ancol là $C_2H_6O$, $C_3H_8O$
b,
$n_{\rm ancol}=\dfrac{0,09}{n}=0,035(mol)=n_{H_2O}-n_{CO_2}$
$\to n_{H_2O}=0,035+0,09=0,125(mol)$
$\to m=0,125.18=2,25g$
Câu 2:
Đặt CTTQ hai ancol là $C_nH_{2n+2}O$
$n_{H_2}=\dfrac{2,688}{22,4}=0,12(mol)$
$\to n_{\rm ancol}=n_{OH}=2n_{H_2}=0,12.2=0,24(mol)$
$\to \overline{M}_{\rm ancol}=\dfrac{13,3}{0,24}=55,4$
$\to 14n+18=55,4$
$\to n=2,67$
Vậy CTPT hai ancol là $C_2H_6O, C_3H_8O$