$n_{O_2}=\dfrac{20,16}{22,4}=0,9(mol)$
Đặt $4x$, $3x$ là số mol $CO_2$, $H_2O$
BTKL:
$44.4x+18.3x=0,9.32+17,2$
$\Leftrightarrow x=0,2$
$n_C=n_{CO_2}=4x=0,8(mol)$
$n_H=2n_{H_2O}=6x=1,2(mol)$
$\Rightarrow n_O=\dfrac{17,2-0,8.12-1,2}{16}=0,4(mol)$
$n_C : n_H: n_O=0,8:1,2:0,4=2:3:1$
$\Rightarrow$ CTĐGN: $C_2H_3O$
$\Rightarrow$ CTPT: $(C_2H_3O)_n$
$n=1\Rightarrow C_2H_3O$ (loại, không có CTCT)
$n=2\Rightarrow C_4H_6O_2$ (TM)
Vậy A là $C_4H_6O_2$
PTHH:
$C_4H_6O_2+\dfrac{9}{2}O_2\buildrel{{t^o}}\over\to 4CO_2+3H_2O$
