\(n_{Br_2}=\dfrac{28}{160}=0,175\left(mol\right)\\ a,C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ b,n_{C_2H_4}=n_{Br_2}=0,175\left(mol\right)\\ V_{C_2H_4}=0,175.22,4=3,92\left(l\right)\\ \%V_{C_2H_4}=\dfrac{3,92}{8,6}.100\approx45,581\%\\ \%V_{CH_4}\approx54.419\%\)
c, Thiếu dữ kiện