$ n_{Br_2} = \dfrac{m}{M} = \dfrac{8}{160} = 0,05 (mol) \\ C_2H_4 + Br_2 \rightarrow C_2H_4Br_2 \\ n_{C_2H_4} = n_{Br_2} = 0,05 (mol) \\ m_{C_2H_4} = n . M = 0,05 . 28 = 1,4 (g) \\ \%m_{C_2H_4} = \dfrac{m_{C_2H_4}}{m_{hh}}.100\% = \dfrac{1,4}{2,9}.100 = 48,28 \% \\ \%m_{C_2H_6} = 100\% - 48,28\% = 51,72 \% $