$\rm n_{X}=\dfrac{12,395}{24,79}=0,5(mol)$ 

Quy tắc đường chéo 

$\rm {M_{CO_2}}\hspace{5cm}{0,4}\\\hspace{1,4cm}\huge{\searrow}\hspace{2,5cm}\huge{\nearrow}\\\hspace{3,2cm}{16,4}\\\hspace{1,4cm}\huge{\nearrow}\hspace{2,5cm}\huge{\searrow}\\{M_{CH_4}}\hspace{5cm}{27,6}$

`->` $\rm \dfrac{n_{CO_2}}{n_{CH_4}}=\dfrac{0,4}{27,6}=\dfrac{1}{69}$ 

Suy ra $\rm {CO_2}=\dfrac{1}{70}n_X=\dfrac{1}{70}.0,5=\dfrac{1}{140}(mol)$ 

`=>` $\rm n_{CH_4}=0,5-\dfrac{1}{140}=\dfrac{69}{140}(mol)$ 

`d_(X//H_2)=M_X/2=8,2=>M_X=16,4`.

`(n_(CO_2)*44+n_(CH_4)*16)/(n_(CO_2)+n_(CH_4))=16,4`.

`n_(CO_2)*44+n_(CH_4)*16=16,4*n_(CO_2)+16,4*n_(CH_4)`.

`27,6*n_(CO_2)=0,4*n_(CH_4)`.

`n_(CO_2):n_(CH_4)=1/69`. 

`n_(CO_2)=(n_(CH_4))/69(mol)`.

`n_(CO_2)+n_(CH_4)=12,395:24,79=0,5mol`.

`(n_(CH_4))/69+n_(CH_4)=0,5`.

`n_(CH_4)=0,4928571429mol`.

`n_(CO_2)=0,0071428571mol`.