a,
$C_2H_4+Br_2\to BrCH_2-CH_2Br$
b,
$m_{\text{tăng}}=2,8g=m_{C_2H_4}$
$\to n_{C_2H_4}=\dfrac{2,8}{28}=0,1(mol)$
$n_{hh}=\dfrac{6,5}{22,4}=0,29(mol)$
$\to n_{CH_4}=0,29-0,1=0,19(mol)$
$\%m_{CH_4}=\dfrac{0,19.16.100}{0,19.16+0,1.28}=52,05\%$
$\to \%m_{C_2H_4}=47,95\%$