\(n_Y=1\left(mol\right)\)
\(n_{CH_4}=a\left(mol\right)\Rightarrow n_{N_2}=1-a\left(mol\right)\)
\(\overline{M}=\dfrac{16a+28\cdot\left(1-a\right)}{1}=2\cdot12.5=25\left(g\text{/}mol\right)\)
\(\Rightarrow a=0.25\)
\(\%CH_4=\dfrac{0.25}{1}\cdot100\%=25\%\)
\(B\)