%H2=100%-22%-64%=14%
- Chọn 1 mol hỗn hợp. Ta có:
\(n_{CO_2}=\dfrac{22.1}{100}=0,22mol\)
\(n_{O_2}=\dfrac{64.1}{100}=0,64\)
\(n_{H_2}=\dfrac{14.1}{100}=0,14mol\)
\(\overline{M}=\dfrac{0,22.44+0,64.32+0,14.2}{1}=30,44\)
\(d_{\dfrac{X}{H_2}}=\dfrac{30,44}{2}=15,22\)