`n_{khí} = (3,09875)/(24,79) = 0,125` mol

`n_{C_2H_4Br_2} = (4,7)/(188) = 0,025` mol

`C_2H_4 + Br_2 -> C_2H_4Br_2`

Theo PT:

`n_{C_2H_4} = n_{C_2H_4Br_2} = 0,025` mol

`=> %V_{C_2H_4} = (0,025)/(0,125).100% = 20%`

`%V_{CH_4} = 100%-20%=80%`