URGENT CuSO4*5H2O + (NH4)2S2O8?????

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eleaah
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URGENT CuSO4*5H2O + (NH4)2S2O8?????

Post by eleaah »

I can't seem to figure out a reaction for CuSO4*5H2O + (NH4)2S2O8 in the presence of pyridine to make [Cu(py)4]S2O8. Where should the waters go? Should the SO4 join with the ammonium?

Same as for AgNO3 and (NH4)2S2O8 in the presence of pyridine to make [Ag(py)4]S2O8.

Please, please, pleaaase help me out if you're up. I'll probably be up for another 3-4 hours.
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ChenBeier
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Post by ChenBeier »

In aqueous solution you have ions present, what means Cu2+, NH4+ SO4 2-, S2O8 2-

Theoretically you can write Cu2+ + 4 py + S2O8 2- => Cu(py)4S2O8

Ammonium and Sulfate is remaining.

CuSO4*5H2O + (NH4)2S2O8 + 4 py => Cu(py)4S2O8 + (NH4)2SO4 + 5 H2O

For silver its similar.

2 AgNO3 + (NH4)2S2O8 +4 py => Ag2(Py)4S2O8 + 2 NH4NO3
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