$n_{CO_2}=\dfrac{22,4}{22,4}=1(mol)$
Đặt CTTQ ancol là $C_nH_{2n+2}O$
Bảo toàn $C$: $n_A=\dfrac{n_{CO_2}}{n}=\dfrac{1}{n}(mol)$
$\to M_A=\dfrac{20n}{1}=20n=14n+18$
$\to n=3$
Vậy CTPT ancol là $C_3H_8O$
CTCT:
$CH_3-CH_2-CH_2OH$: ancol propylic
$CH_3-CHOH-CH_3$: ancol isopropylic