Question

Care este raportul molar?

Answer 1

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