$a,PTPƯ:C_2H_4+Br_2\xrightarrow{} C_2H_4Br_2$
$b,n_{Br_2}=\dfrac{48}{160}=0,3mol.$
$n_{hỗn\ hợp}=\dfrac{8,96}{22,4}=0,4mol.$
$⇒n_{CH_4}=n_{hỗn\ hợp}-n_{C_2H_4}=0,4-0,3=0,1mol.$
$⇒m_{CH_4}=0,1.16=1,6g.$
$⇒m_{C_2H_4}=0,3.28=8,4g.$
$⇒\%m_{CH_4}=\dfrac{1,6}{1,6+8,4}.100\%=16\%$
$⇒\%m_{C_2H_4}=\dfrac{8,4}{1,6+8,4}.100\%=84\%$
chúc bạn học tốt!