Đáp án:
$C_{2}H_{4} + Br_{2} → C_{2}H_{4}Br_{2}$
x x
$C_{2}H_{2} + 2Br_{2} → C_{2}H_{4}$
y 2y
$nBr_{2}=\frac{100,8}{160}=0,63$
$nKhí=\frac{10,08}{22,4}=0,45$
$x+y=0,45$
$x+2y=0,63$
⇒$\left \{ {{x=0,27} \atop {y=0,18}} \right.$
$V\%C_{2}H_{4}=\%nC_{2}H_{4}=\frac{0,27}{0,45}.100=60\%$
$V\%C_{2}H_{2}=100-60=40\%$