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Topic: Phosphate/ammonium molybdate  (Read 19171 times)

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Offline dahoog

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Phosphate/ammonium molybdate
« on: May 06, 2010, 02:42:59 PM »
Could someone help me in fully balancing the reaction involving Phosphate and ammonium molybdate.
One book gave me this 
7PO3- + 12(NH4)6Mo7O24 + 36H2O → 7(NH4)3PO4 ∙ 12Mo + 51NH4+ + 72 OH-
Yet I am after the reaction where [PMo12O40]3- is created and PO4 ∙ 12Mo alone doesn't seem right.
Any help would be greatly appreciated.

Offline Borek

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Re: Phosphate/ammonium molybdate
« Reply #1 on: May 06, 2010, 04:05:58 PM »
Try

H3PO4 + (NH4)2MoO4 + HNO3 -> (NH4)3PO4·12MoO3 + NH4NO3 + H2O
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Offline dahoog

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Re: Phosphate/ammonium molybdate
« Reply #2 on: May 09, 2010, 12:43:18 AM »
Is PO4·12MoO3 the same as [PMo12O40]3-?

Offline Borek

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Re: Phosphate/ammonium molybdate
« Reply #3 on: May 09, 2010, 03:47:54 AM »
It is just a matter of notation.

Beware: you need to keep charges correct.
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Offline dahoog

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Re: Phosphate/ammonium molybdate
« Reply #4 on: May 15, 2010, 12:21:36 AM »
For future reference

7[PO4]3- + 12[Mo7O24]6- + 72H+→ 7[PMo12O40]3- + 36H2O

Offline larryp7639

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Re: Phosphate/ammonium molybdate
« Reply #5 on: May 21, 2010, 04:25:34 AM »
For future reference

7[PO4]3- + 12[Mo7O24]6- + 72H+→ 7[PMo12O40]3- + 36H2O


Thanks you for the post.

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