a,
$n_{CO_2}=\dfrac{1,12}{22,4}=0,05(mol)$
Đặt CTTQ ankadien X là $C_nH_{2n-2}$
Bảo toàn $C$:
$n.n_X=n_{CO_2}$
$\to n_X=\dfrac{0,05}{n}(mol)$
$\to M_X=\dfrac{0,68n}{0,05}=13,6n=14n-2$
$\Leftrightarrow n=5$
Vậy CTPT X là $C_5H_8$
b,
CTCT ankadien X:
$CH_2=C=CH-CH_2-CH_3$
$CH_2=CH-CH=CH-CH_3$
$CH_2=CH-CH_2-CH=CH_2$
$CH_3-CH=C=CH-CH_3$
$CH_2=C(CH_3)-CH=CH_2$
$CH_3-C(CH_3)=C=CH_2$
