$V_{O_2}=3,22.20\%=0,644(l)$
Gọi CTTQ 2 ankan là $C_nH_{2n+2}$
$C_nH_{2n+2}+\dfrac{3n+1}O_2\buildrel{{t^o}}\over\to nCO_2+(n+1)H_2O$
$V_{\text{ankan}}=0,28(l)$
$\to V_{O_2}=\dfrac{3n+1}{2}.0,28=0,14(3n+1)(l)$
$\to 0,14(3n+1)=0,644$
$\to n=1,2$
Vậy CTPT 2 ankan là $CH_4, C_2H_6$