\(m_{bình.tăng}=m_{C2H4}=\dfrac{4,48}{28}=0,16\left(mol\right)\)
a) Pt : \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b) \(\left\{{}\begin{matrix}\%V_{C2H4}=\dfrac{0,16.24,79}{8,9244}.100\%=44,44\%\\\%V_{CH4}=100\%-44,44\%=55,56\%\end{matrix}\right.\)
c) Theo Pt : \(n_{C2H4}=n_{Br2}=0,16\left(mol\right)\)
\(\rightarrow V_{ddBr2}=\dfrac{0,16}{1,6}=0,1\left(l\right)\)