$C_nH_{2n-6}+(1,5n-1,5)O_2\rightarrow n{CO_2}+(n-3)H_2O$$\Rightarrow $Giả sử ban đầu $\begin{cases}n_X=1 mol \\n_{O_2}=19 mol \end{cases}\Rightarrow n_{trước}=20 mol\Rightarrow n_{sau}=\frac{20.1,05}{1}=21 mol$
$\Rightarrow [n+(n-3)]-[1+(1,5n-1,5)]=21-20\Rightarrow n=7\Rightarrow C_6H_5-CH_3$