Câu 1:
$n_{CO_2}=0,2 mol$
$n_{H_2O}=0,2 mol= n_{CO_2}$
$\Rightarrow$ Axit no, đơn chức.
$C_nH_{2n}O_2+\dfrac{3n-1}{2}O_2\to nCO_2+nH_2O$
$\Rightarrow n_X=\dfrac{0,2}{n}$
$\Rightarrow M_X=\dfrac{4,4n}{0,2}=22n=14n+32$
$\Leftrightarrow n=4$ ($C_4H_8O_2$)
CTCT:
(1) $CH_3-CH_2-CH_2-COOH$ (axit butanoic)
(2) $CH_3-CH(CH_3)-COOH$ (axit 2-metylpropanoic)
Câu 2:
$n_{H_2}=\dfrac{7,392}{22,4}=0,33 mol$
$\Rightarrow n_{C_6H_5OH}=2n_{H_2}=0,66 mol$
$m=0,66.94=62,04g$