$C_nH_{2n+2}O + \dfrac{3n}{2}O_2\xrightarrow{t^o} nCO_2 + (n+1)H_2O$
0,2 : 0,3 (mol)
$n_{CO_2}=\dfrac{4,48}{22,4}=0,2(mol)$
$n_{H_2O}=\dfrac{5,4}{18}=0,3(mol)$
Ta có: $0,2.(n+1)=0,3n$
<=>$n=2$
=> X là $C_2H_6O$
CTCT: $CH_3CH_2OH$
Danh pháp thay thế: ancol etylic
