Đặt `n_{CH_4}=x(mol);n_{C_2H_4}=y(mol);n_{H_2}=z(mol)`
`->x+y+z={3,36}/{22,4}=0,15(1)`
`C_2H_4+Br_2->C_2H_4Br_2`
`->m_{C_2H_4}=m_{\text{bình tăng}}=0,84(g)`
`->28y=0,84`
`->y=0,03(2)`
`M_A={0,975}/{{1,4}/{22,4}}=15,6(g//mol)`
`->m_A=15,6.0,15=2,34=16x+28y+2z(3)`
`(1)(2)(3)->x=0,09(mol);y=z=0,03(mol)`
Vậy trong hỗn hợp đầu:
`V_{H_2}=V_{C_2H_4}=0,03.22,4=0,672(l)`
`V_{CH_4}=0,09.22,4=2,016(l)`