a) PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Hiện tượng: Dung dịch Brom bị nhạt màu
b) Ta có: \(n_{Br_2}=\dfrac{24}{160}=0,15\left(mol\right)=n_{C_2H_4Br_2}\)
\(\Rightarrow m_{C_2H_4Br_2}=0,15\cdot188=28,2\left(g\right)\)
c) Theo PTHH: \(n_{C_2H_4}=0,15\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,15\cdot22,4}{4,48}\cdot100\%=75\%\)
\(\Rightarrow\%V_{CH_4}=25\%\)