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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: dahoog on May 06, 2010, 02:42:59 PM

Title: Phosphate/ammonium molybdate
Post by: dahoog 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.

Title: Re: Phosphate/ammonium molybdate
Post by: Borek on May 06, 2010, 04:05:58 PM

Try

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

Title: Re: Phosphate/ammonium molybdate
Post by: dahoog on May 09, 2010, 12:43:18 AM

Is PO4·12MoO3 the same as [PMo12O40]3-?

Title: Re: Phosphate/ammonium molybdate
Post by: Borek on May 09, 2010, 03:47:54 AM

It is just a matter of notation.

Beware: you need to keep charges correct.

Title: Re: Phosphate/ammonium molybdate
Post by: dahoog on May 15, 2010, 12:21:36 AM

For future reference

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

Title: Re: Phosphate/ammonium molybdate
Post by: larryp7639 on May 21, 2010, 04:25:34 AM

Quote from: dahoog on May 15, 2010, 12:21:36 AM

For future reference

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

Thanks you for the post.

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