$2C_xH_y+\dfrac{4x+y}{2}O_2\to 2xCO_2+yH_2O>
$n_A=1 \Rightarrow n_{O_2}=0,25(4x+y)$
$\Rightarrow 0,25(4x+y)=6$
$\Leftrightarrow 4x+y=24$
$\Rightarrow x=4; y=8$ $(C_4H_8)$
CTCT:
(1)$ CH_2=CH-CH_2-CH_3$
(2) $CH_3-CH=CH-CH_3$
(3) $CH_2=C(CH_3)-CH_3$
(Và thêm 2 CTCT mạch vòng no 4 cạnh, 3 cạnh)