a) \(n_{Br_2}=\dfrac{4}{160}=0,025\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,025<-0,025
=> VC2H4 = 0,025.22,4 = 0,56 (l)
=> VCH4 = 5,6 - 0,56 = 5,04 (l)
b) \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{5,04}{5,6}.100\%=90\%\\\%V_{C_2H_4}=\dfrac{0,56}{5,6}.100\%=10\%\end{matrix}\right.\)


