Bibinou Posted March 21, 2006 Posted March 21, 2006 Good morning, I have problems in balancing a synthesis reaction. The pathway is given : (NH4)6Mo7024.4H2O +28 H2S+8 NH4OH--> 7(NH4)2MoS4 + 36 H20 NH4)2MoS4 (CuCN/KCN/[Pr4N)Br]) --> (Pr4N)2[(CN)CuS2MoS2] (Pr4N)2[(CN)CuS2MoS2] + Cu(PPh3)2(NO3) --> [(Ph3P)2CuS2MoS2Cu(PPh3)] I don't really see how to equilibrate the 2 last ones and I have also problems to explain really what occurs during this synthesis. How do we explain the fact that we use KCN AND CuCN? Could we also form KBr and NH4CN for the second reaction? I found on a contribution from the department of chemistry in Australia. The follwing paragraph explains how works the second reaction : Synthesis of (n-Pr4N)2[(CN)CuS2MoS2] a solution of KCN (O.64g 9.8mmol) in H2O (7 cm3) is added to a suspension of NH4CuMoS4 (3g 9.8mmol) in H2O (30 cm3) to produce a deep red solution. A solution of (n-Pr4N)Br (5.2g 19.6 mmol) in H20 (10cm3) is added dropwise producing a red precipitate. The mixture is stirred 15min before the compound which is washed with H2O, EtOH and ether and dried in air, is isolated. (yield 65% 4.35g) Ph4As+ et Et4N+ may be prepared in a similar manner. The compounds may also be prepared by combinating equimolar aqueous solution of (NH4)2[MoS4] and [Cu(CN)x ]^(1-x) (produced by combinatingCuSO4.5H20 orCuCN with KCN) followed by the addition of 2-3 equivalents of the quaternary cation. Thank you very much for your help.
woelen Posted March 21, 2006 Posted March 21, 2006 In the last equation, where is the CN-group going on the right side? The same for the nitrate. I think there is incomplete information to solve this. These equations cannot be balanced.
Bibinou Posted March 22, 2006 Author Posted March 22, 2006 I know that there no CN in the last equation In fact I don't know what is produced trough this synthesis and and hoped that somebody would have a brilliant idea...
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