a/
\(\%V_{H_2}=60\%\rightarrow\%V_{CH_4}=40\%\)
\(\%m_{H_2}=12\%\rightarrow\%m_{CH_4}=88\%\)
b/
xét 1 mol hỗn hợp B:
ta có:\(n_{H_2}=0,6\left(mol\right)\left(60\%\right)\)\(\rightarrow m_{H_2}=0,6.2=1,2\left(g\right)\)
\(n_{CH_4}=0,4\left(mol\right)\left(40\%\right)\rightarrow m_{CH_4}=16.0,4=6,4\left(g\right)\)
\(\Rightarrow m_B=6,4+1,2=7,6\left(g\right)\)
xét 1 mol \(H_2\) :\(m_{H_2}=1.2=2\left(g\right)\)
\(D_{\dfrac{B}{H_2}}=\dfrac{7,6}{2}=3,8\)