67,2/22,4 = 3 moli amestec propina + 1,3-butadiena
solutia de Br2, avem
Vs = 3 L, Cm = 2 M , stim ca Cm = niu/Vs
=> niu = CmxVs = 2x3 = 6 moli
a moli an moli
C3H4 + nBr2 --CCl4--> C3H4Br2n
1 n
b moli bm moli
C4H6 + mBr2 --CCl4--> C4H6Br2m
1 m
an + bm = 6 moli (-)
a + b = 3 moli
----------------------------
an-a + bm-b = 3 moli
=> a(n-1) + b(m-1) = 3
a/b = an/bm => n/m = 1
=> n = m => a = b => 1:1
