$a,PTPƯ:C_2H_4+Br_2\xrightarrow{} C_2H_4Br_2$
$b,n_{C_2H_4Br_2}=\dfrac{9,4}{188}=0,05mol.$
$Theo$ $pt:$ $n_{Br_2}=n_{C_2H_4}=n_{C_2H_4Br_2}=0,05mol.$
$⇒m_{Br_2}=0,05.160=8g.$
$c,\%V_{C_2H_4}=\dfrac{0,05.22,4}{2,4}.100\%=46,67\%$
$\%V_{CH_4}=100\%-46,67\%=53,33\%$
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