a) \(n_{C_2H_4}=\dfrac{0,112}{22,4}=0,005\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,005-->0,005
=> \(V_{dd.Br_2}=\dfrac{0,005}{0,1}=0,05\left(l\right)=50\left(ml\right)\)
b) \(n_{C_2H_2}=\dfrac{0,112}{22,4}=0,005\left(mol\right)\)
PTHH: C2H2 + 2Br2 --> C2H2Br4
0,005->0,01
=> \(V_{dd.Br_2}=\dfrac{0,01}{0,1}=0,1\left(l\right)=100\left(ml\right)\)