$C_nH_{2n-6}+\frac{3n-3}{2}O_2\Rightarrow nCO_2+(n-3)H_2O$
Ta có: $nO_2=1,3125 mol, nC_nH_{2n-6}=\frac{13,25}{14n-6}$
- Từ phương trình, ta có: $nO_2=\frac{3n-3}{2}nC_nH_{2n-6}\Leftrightarrow \frac{13,25}{14n-6}.\frac{3n-3}{2}=1,3125\Leftrightarrow n=8\Rightarrow C_8H_{10}.$