\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Ta có: \(n_{C2H4}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Theo phản ứng: \(n_{Br2}=n_{C2H4}=0,1\left(mol\right)\rightarrow m_{Br2}=0,1.160=16\left(g\right)\)
\(\rightarrow m_{dd_{brom}}=\frac{16}{20\%}=80\left(g\right)\)