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

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

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