a) \(C_2H_4+Br_2\xrightarrow[]{}C_2H_4Br_2\)
b) \(n_{C_2H_4Br_2}=n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(m_{C_2H_4Br_2}=0,5.188=94\left(g\right)\)
c) \(n_{Br_2}=n_{C_2H_4}=0,5\left(mol\right)\)
500ml=0,5l
\(C_{M_{Br_2}}=\dfrac{0,5}{0,5}=1\left(M\right)\)