\(a) C_2H_4 + Br_2 \to C_2H_4Br_2\\ b) m_{C_2H_4} = m_{tăng} = 2,8(gam)\\ \Rightarrow n_{C_2H_4} = \dfrac{2,8}{28} = 0,1(mol)\\ \Rightarrow n_{CH_4} = \dfrac{6,5-0,1.22,4}{22,4} = \dfrac{213}{1120}(mol)\\ \%m_{C_2H_4} = \dfrac{2,8}{2,8 + \dfrac{213}{1120}.16}.100\% = 47,92\%\\ \%m_{CH_4} = 100\% -47,92\% = 52,08\%\)