Ta có:
\(n_{hh}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{C2H4Br2}=n_{Br2}=n_{C2H4}=\frac{18,8}{188}=0,1\left(mol\right)\)
\(\left\{{}\begin{matrix}V\%_{C2H4}=\frac{0,1}{0,25}.100\%=40\%\\V\%_{CH4}=100\%-40\%=60\%\end{matrix}\right.\)
Đổi 200ml = 0,2l
\(\Rightarrow CM_{Br2}=\frac{0,1}{0,2}=0,5M\)
C2H4+Br2->C2H4Br2
0,1---0,1------0,1
ta có nC2H4Br2=18,8\188=0,1 mol
=>VC2H4=0,1.22,4=2,24l
=>VCH4=5,6-2,24=3,36l
=>%C2H4=2,24\5,6.100=40%
=>%CH4=60%
=>CMddBr2=0,5M