a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b, \(n_{C_2H_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Br_2}=n_{C_2H_4Br_2}=n_{C_2H_4}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{Br_2}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
c, \(m_{C_2H_4Br_2}=0,1.188=18,8\left(g\right)\)