Đặt nCH4 = x mol; nC2H4 = y mol.
Ta có hpt: {x + y = 0,15
2x + 2y = 0,2
=> {x = 0,05
..........y = 0,05
→ nX = x + y = 0,05 + 0,05 = 0,1 mol → V = 0,1 x 22,4 = 2,24 lít
Đặt \(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_4}=y\left(mol\right)\end{matrix}\right.\)
Ta có hpt: \(\left\{{}\begin{matrix}x+y=0,15\\2x+2y=0,2\end{matrix}\right.\)
Giải ra: \(\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)
\(\Rightarrow n_X=x+y=0,05+0,05=0,1\left(mol\right)\)
\(\Rightarrow V=0,1.22,4=2,24\left(l\right)\)
Vậy ...............
Đặt \(n_{CH_4}=x\left(mol\right)\) ; \(n_{C_2H_4}=y\left(mol\right)\)
Ta có hpt: \(\left\{{}\begin{matrix}x+y=0,15\\2x+2y=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)
\(\Rightarrow n_X=x+y=0,05+0,05=0,1\left(mol\right)\)
\(\rightarrow V=0,1.22,4=2,24\left(l\right)\)