$\overline{M}_{hh}=14,25.2=28,5$
$n_{\text{ankan}}=\dfrac{0,448}{22,4}=0,02(mol)$
$n_A=\dfrac{1,792}{22,4}=0,08(mol)$
$\to n_{\text{anken}}=0,08-0,02=0,06(mol)$
$m_{\text{anken}}=1,96g$
Đặt CTTQ 2 anken là $C_{\overline{n}}H_{2\overline{n}}$
$\to \overline{M}_{\text{anken}}=\dfrac{1,96}{0,06}=\dfrac{93}{8}=14\overline{n}$
$\to \overline{n}=2,33$
Vậy CTPT 2 anken là $C_2H_4$, $C_3H_6$
$m_A=0,08.28,5=2,28g$
$\to m_{\text{ankan}}=2,28-1,96=0,32g$
Đặt CTTQ ankan là $C_nH_{2n+2}$
$M_{C_nH_{2n+2}}=\dfrac{0,32}{0,02}=16=14n+2$
$\to n=1$
Vậy CTPT anken là $CH_4$
$\%V_{CH_4}=\dfrac{0,448.100}{1,792}=25\%$