# chemistry

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write the equation for copper(II) sulfate pentahydrate and ammonium sulfate forming:
Cu(NH4)2(SO4)2*6H2O

not sure how to balance it

Check your question very carefully. Is this a question given to you by a teacher?

yeah it is in a lab

CuSO4.5H2O + (NH4)2SO4 >>Cu(NH4)2(SO4)2*6H2O

Note a water is missing on the left side, if you really need 6H2O on the right, you will have to add H2O to the left.

so would this be a redox reaction

What do you think is oxidized and reduced?

No. Where did you get that idea?

the product would be oxidized?

I don't see any element that has changed oxidation state.
Cu is +2 on the left and right.
S is +6 on left and right.
O is -2 on left and right.
N is -3 on left and right.

I think it is only fair to point out that I have taught chemistry for almost 50 years and I never heard of Cu(NH4)2(SO4)2. That doesn't mean that it doesn't exist and I may have overlooked something in your question (or I may have overlooke something in these last several years). But I hope we have helped you. Thanks for using the Jiska help site.

i really need a suggestion here.

NH3 + CuSO4*5H2O = Cu(NH3)4SO4*H2O

You need to be more specific about the kind of suggestion for which you are looking. And yes, CuSO4*5H2O + NH3 will produce Cu(NH3)4SO4 + H2O but it won't do it with (NH4)2SO4. Sometimes (NH4)2SO4 is added to CuSO4 BEFORE NH3 is added in order to keep Cu(OH)2 from forming upon the initial addition of NH3.

oh thanks!
so how would i balance the second equation i sent iN?

I don't undersand why we are playing "twenty questions" here. You could tell us what you did at the get go? As for balancing, surely you can balance that last equation. However, IF that is the equation you want balanced, a hint is to add 4NH3. You can take it from there.

i am wondering what reasonable values would be for the x in the formula [Cu(NH3)x]SO4*H2O