\(\begin{array}{l}
n_{O_2}=\frac{20,8}{32}=0,65(mol)\\
A:\,C_nH_{2n+2}\\
C_nH_{2n+2}+\frac{3n+1}{2}O_2\xrightarrow{t^o}nCO_2+(n+1)H_2O\\
Theo\,PT:\,n_{C_nH_{2n+2}}.\frac{3n+1}{2}=n_{O_2}\\
\to \frac{5,8(3n+1)}{2(14n+2)}=0,65\\
\to n=4\\
\to A:C_4H_{10}
\end{array}\)