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How do you balance H3PO4 + (NH4)*2MoO4 + HNO3 --> (NH4)3PO4*12MoO3 + NH4NO3 + H2O?
This isn't a homework question! I'm prepping for an exam, and I can't figure this out. I know what the answer is supposed to be: H3PO4 + 12(NH4)*2MoO4 + 21HNO3 --> (NH4)3PO4*12MoO3 + 21NH4NO3 + 12H2O. Could you please give me the steps on how to get there? Thanks!
1 Answer
- kumorifoxLv 71 decade agoFavorite Answer
This is not a redox, just a complicated rearrangement, so balancing is actually simpler than you'd think. I'll do it step-wise, to show you what happens. Remember to leave H and O until last; do the more unusual elements first, then keep the groups together and do each side in turn.
H3PO4 + (NH4)2MoO4 + HNO3 → (NH4)3PO4∙12MoO3 + NH4NO3 + H2O
H3PO4 + 12(NH4)2MoO4 + HNO3 → (NH4)3PO4∙12MoO3 + NH4NO3 + H2O (balance the Mo)
H3PO4 + 12(NH4)2MoO4 + HNO3 → (NH4)3PO4∙12MoO3 + 21NH4NO3 + H2O (balance the NH4)
H3PO4 + 12(NH4)2MoO4 + 21HNO3 → (NH4)3PO4∙12MoO3 + 21NH4NO3 + H2O (balance the NO3)
H3PO4 + 12(NH4)2MoO4 + 21HNO3 → (NH4)3PO4∙12MoO3 + 21NH4NO3 + 12H2O (balance the H and O; the H comes from the acids)
This is now balanced.
