a)
\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4Br_2} = n_{C_2H_4} = n_{Br_2} = \dfrac{160.5\%}{160} = 0,05(mol)\\ \Rightarrow m_{C_2H_4Br_2} = 0,05.188 = 9,4(gam)\)
b)
\(\%V_{C_2H_4} = \dfrac{0,05.22,4}{4,48}.100\% = 25\%\\ \%V_{CH_4} = 100\% - 25\% = 75\%\)