Chemistry

posted by .

My question is basically to find out what these reactants yield:

CuCl2(aq) + NH3(aq) --> ?

my best guess here is...
CuCl2(aq) + 6NH3(aq)-->[Cu(NH3)4(H2O)2]2+ + N2(g) + 2HCl

Im pretty sure about the [Cu(NH3)4(H2O)2]+2 part but not sure about the rest.

The primary product of this reaction is the copper(II) ammine complex ion, Cu(NH3)42+ and you make the remainder balance. I believe the ammine complex has a coordination number of 4 and not 6 although I'm not up on the latest research. I would do it like this.
CuCl2(aq) + 6NH3(aq) + 2HOH(l) ==>Cu(NH3)4^+2(aq) + 2Cl^-(aq) + 2NH4^+(aq) + 2 OH^-(aq).
There is absolutely no reason to believe N2(g) will be formed (and it isn't). If I find something to indicate a coordination number of six I will post another note here.

I later reviewed the work the reaction that is done is a series of copper related reactions, in a flow chart given it shows that the reactants CuCl2 and NH3 yield a product [Cu(NH3)4(H20)2]2+(aq) complex but i was wondering that clorine ions are present in the solution but how are the remaining nitrogen ions and hydrogen ions? are they floating around in this solution?

Any help I greatly appreciate.

I searched on the Internet and found what I believe to be a reliable source (from the UK) that confirms the coordination number of 6 with four NH3 and 2H2O; therefore, the complex ion [Cu(NH3)4(H2O)2]^2+ is entirely appropriate. The CuCl2 is present in aqueous solution as Cu^+2 and Cl^- and both of those stay as ions through the process; therefore, the Cl^- appears on the product side. There are no nirogen ions and few, if any, H^+ (especially in an NH3 solution). Actually, there are no Cu^+2 ions at the beginning either. Since this is an aqueous solution, I am sure the Cu^+2 is present as [Cu(H2O)6]^+2. If we start with that the equation is easier to write.
[Cu(H2O)6]Cl2 + 4NH3 ==>[Cu(NH3)4(H2O)2]^+2 + 2Cl^- + 4H2O.

Getting back to the original equation, if we write it like this,
CuCl2(aq) + 4NH3(aq) + 2H2O(l) ==> [Cu(NH3)4(H2O)2]^+2(aq) + 2Cl^-(aq)
we stick closer to what you are to start with and end with. Everything balances and you have no "nitrogen ions" to be concerned about.
Technically, I don't like to see NH3(aq) in solution without noting that it forms NH3 + HOH ==> NH4^+ + OH^-. IF you had H^+ that you asked about they would be neutralized with OH^- from the NH3 reaction. None of the ammonia will dissociate into a "nitrogen ion." I guess the short answer to your question, " but i was wondering that clorine ions are present in the solution but how are the remaining nitrogen ions and hydrogen ions? are they floating around in this solution?" is
CuCl2(aq) + 4NH3(aq) + 2H2O)l)==>[Cu(NH3)4(H2O)2]^+2(aq) + 2Cl^-(aq), which is the equation I wrote above, has no hydrogen ions or nitrogen ions to be conerned about.


Thanks for using the Jiskha boards. I hope this clears up any misunderstandings. Please follow up with anything you don't understand.



will it be
NH3 (aq) = NH4OH

NH4OH + CuCl2 --> Cu(NH3)Cl2 + H2O ?

I'm doing a lab on equilibriums with this equation, well actually we have to find the equation, but he did say that [Cu(H2O)6]+2 and [Cu(NH3)4(H20)2]+2 would be involved like you said, but some where in between these two he said there is some Cu(OH)2, which results from adding a little NH4OH to the CuCl2, uppon addin't more though it finally becomes what you had as your product, any idea what that middle step might be?

  • Eitoku High School -

    I think the answer is just simply like this:
    NH3(aq) --> NH4OH
    NH4OH + CuCl2 --> Cu(OH)2 + NH4Cl

  • Chemistry -

    well this sucks. i have to write a lab report on this. i know the above reacton forms a blue soilution, and cu(no3)2 is blue in water...hmmm...

  • Chemistry -

    NH3(aq) --> NH4OH
    NH4OH + CuCl2 --> Cu(OH)2 + NH4Cl
    First a massive precipitate forms (the copper hydroxide), however as more ammonia is added the equilibrium shifts
    to the diaquatetrammine copper(II) complex.
    Your best guess:
    CuCl2(aq) + 6NH3(aq) -->[Cu(NH3)4(H2O)2]2+ + N2(g) + 2HCl

    I'd think that this is a more probable reaction:
    CuCl2(aq) + 4NH3(aq)-->[Cu(NH3)4(H2O)2]2+ Cl2 , try smelling the products :)
    (if you have the means to make them)

  • Chemistry -

    There is no answer because both ionic compounds are aqeous or soluble.

  • Chemistry -

    :D lol
    CuCl2 + 4NH3 --> [Cu(NH3)4]Cl2

  • Chemistry -

    already balanced

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chemistry

    from the balanced equation 4NH3 +7O2 - 4NO2 + 6H2O How many molecules of water are produced when 2.25 moles of ammonia are completely reacted?
  2. AP Chemistry

    I have almost the same question as Josh from August 23rd. If you use 10.0 g of CuSO4 and excess NH3, what is the theoretical yield of Cu(NH3)4SO4?
  3. Chemistry - please answer

    I think I managed to totally screw up my computer so I did already post this question, but from what I see it is gone now. I don't really know if its actually there or not cuz I can't do anything with computers but here is the question: …
  4. Chemistry

    Write balanced equations for the formation of [Ni(NH3)6][BF4]2 starting from an anhydrous sample of nickel(II) choride hexahydrate. (3 equations necessary) I am slightly confused by this. I think the starting sample has the formula …
  5. Honors Chemistry

    "How many grams of NH3 can be produced from the reaction of 28 g of N2 and 25 g of H2?
  6. college chem

    The cation M2+ reacts with NH3 to form a series of complex ions as follows: M2+ + NH3 M(NH3)2+ K1 = 102 M(NH3)2+ + NH3 M(NH3)2 2+ K2 = 103 M(NH3)2 2+ + NH3 M(NH3)3 2+ K3 = 102 A 1.0 × 10–3 mol sample of M(NO3)2 is added to 1.0 L …
  7. chemistry

    what is the Molarity (M) of a 0.87m aqueous solution of ammonia, NH3?
  8. Chemistry

    1. Ammonia gas can be prepared by the following unbalanced equation: _____ CaO(s) + _____ NH4Cl (s) --> _____ NH3(g) + _____ H2O(g) + _____ CaCl2(s) a. Balance the equation. CaO + 2NH4Cl = 2NH3 + H2O + CaCl2 b. Suppose you begin …
  9. chem

    Calculate the equilibrium concentration of NH3,Cu^2+,[ Cu(NH3)]^2+ ,[Cu(NH3)2]^2+,[Cu(NH3)3]^2+,and[Cu(NH3)4]^2+ in a solution made by mixing 500.0 ml of 3.00 M NH3 with 500.0 ml of 2.00*10^-3 M Cu(NO3)2. The sequential equilibria …
  10. Chemistry

    Order in the increasing value of their polarity?

More Similar Questions