khí không tác dụng là $\rm CH_4$

`=>` $\rm V_{CH_4}=2,24(l)$ $\\$$\rm n_{CH_4}=\dfrac{2,24}{22,4}=0,1(mol)$$\\$$\rm a.V_{CH_4}=2,24(l)$$\\$$\rm V_{C_2H_4}=6,72-2,24=4,48(l)$

`b.%V_{CH_4}={2,24}/{6,72}.100%~~33,33%`

`%V_{C_2H_4}=100-33,33=66,67%`

`C_2H_4 + Br_2 -> C_2H_4Br_2`

Khí thoát ra là `CH_4` không phản ứng.

a)

`V_{CH_4} = 2,24l`

`V_{C_2H_4} = 6,72 - 2,24 = 4,48l`

b)

`%V_{CH_4} = (2,24)/(6,72) . 100% ~~ 33,333%`

`%V_{C_2H_4} = 100% - 33,333% = 66,667%`