$n_{H_2O}=\dfrac{18}{18}=1(mol)$
Đặt CTTQ rượu $A$ là $C_nH_{2n+2}O$
$C_nH_{2n+2}O\xrightarrow{{O_2, t^o}} nCO_2+ (n+1)H_2O$
$\to n_A=\dfrac{1}{n+1}(mol)$
$\to M_A=\dfrac{15}{\dfrac{1}{n+1}}=15(n+1)=15n+15=14n+18$
$\to n=3$
Vậy CTPT $A$ là $C_3H_8O$
$n_A=\dfrac{1}{3+1}=0,25(mol)$
$\to n_{CO_2}=3n_{C_3H_8O}=0,75(mol)$
$\to x=m_{CO_2}=0,75.44=33g$