When aqueous copper(II)chloride reacts with aqueous ammonium phosphate, soluble ammonium chloride forms and copper(II) phosphate precipitates out of solution.
1. Write the balanced molecular equationfor this reaction.
2. Write the balanced complete ionic equation for this reacton.
(I don't understand the difference in the two equations so obviously I can't write them!)
3. What are the spectator ions in this reaction?
(I know what spectator ions are but I don't know how you tell which ions are spectator ions)
4. Write the net ionic equation for this reaction.
I need some major help with this problem!
For Further Reading
Chemical Formulas and Reactions - DrBob222, Friday, December 29, 2006 at 6:48pm
CuCl2 + (NH4)3PO4==>Cu3(PO4)2 + NH4Cl is the molecular equation.
1. Balance the molecular equation. I will leave that for you.
2. Now, separate the soluble materials into ions. Leave insoluble materials as the molecule. Example: NH4Cl will be written as NH4^+ + Cl^- and Cu3(PO4)2 will be written as Cu3(PO4)2. YOu know CuCl2 is soluble and you know (NH4)3PO4 is soluble. When you finish this part you have written the ionic equation.
3. Look through the ionic equation you have completed and find the ions that appear on both sides of the equation. These are the spectator ions; i.e., they don't enter into the reaction.
4. Cancel the spectator ions and you will be left with the NET ionic equation.
Get as far as you can, post your work to the point you get stuck and we shall be happy to help you through the remainder.
Chemical Formulas and Reactions - chrissy, Friday, December 29, 2006 at 8:05pm
Ok so I think I balanced the molecular equation but I'm not sure if it is correct.
1. Cu3Cl+(NH4)2PO6---->Cu(PO4)3+NH4Cl2
then for #2:
Cu3(PO4)2 ---> NH4++Cl-
but then I can't find the spectator ions because there are no ions that appear on both sides of this equation so I think I did it wrong or there are just no spectator ions in this equation.
I haven't found the net ionic equation because I don't think I've done any of this problem right so far. Can you help me figure this out please?!
Chemical Formulas and Reactions - DrBob222, Friday, December 29, 2006 at 10:40pm
The number 1 rule in balancing equations is you may NOT change the subscripts. You balance equations ONLY by changing the coefficients.
CuCl2 + (NH4)3PO4==>Cu3(PO4)2 + NH4Cl
HINT:
3CuCl2 + 2(NH4)3PO4 ==> Cu3(PO4)2 + 6NH4Cl
Try it from here
Ok I am still having difficulty balancing this equation. I don't understand why in the hint you gave me 3 is in front of CuCl2(3CuCl2)and I was wondering where the 6 came from in (6NH4Cl).Can you walk me through how to balance this molecular equation:
CuCl2+(NH4)3PO4--->Cu3(PO4)2+NHCl
I am totally lost and I need to get this paper in.
In the hint I gave you, I BALANCED the equation. It is balanced as I wrote it. Let's check it.
I see 3 Cu atoms on the left (from 3CuCl2) and 3 on the right (from Cu3(PO4)2).
I see 6 Cl atoms on the left (from 3CuCl2) and 6 on the right (from 6NH4Cl).
I see 6NH4^+ ions on the left (from 2(NH4)3PO4) and 6 on the right (from 6NH4Cl).
I see 2 PO4 ions on the left (from 2(NH4)3PO4) and 2 phosphate ions on the right (from Cu3(PO4)2).
so the equation balances as is. The 6 for NH4Cl came because I put it there to balance the equation. I put all of the coefficients in. I did not change any of the subscripts.
Where do we go from here?
I'm sorry I was just trying to figure out how you came up with those numbers I see it now. Ok now I need to find the balanced complete ionic equation for this reaction. I think it is
3Cu(PO4)---> 6(NH4)+Cl- but i'm not to confident in that answer.
To write the balanced ionic equation, we need to separate the materials into their ions. The rules are:
1. Soluble compounds are written as the ions (sometimes with (aq) after it but I don't know if your instructor wants that or not).
2. insoluble compounds are written as the molecule.
CuCl2 is soluble so you write it as Cu^+2(aq) + 2Cl^-. Of course you have a coefficient of 3 for CuCl2; therefore, it would be written as 3Cu^+2(aq) + 6Cl^-(aq).
We had 2(NH4)3PO4. It is soluble; therefore, it becomes 6NH4^+(aq) + 2PO4^-3(aq).
Cu3(PO4)2 is a ppt (insoluble) so it is written as Cu3(PO4)2(s) where the (s) stands for solid.
6NH4Cl is soluble and it is written as 6NH4^+ + 6Cl^-
Write that down on paper so you can look at it, then cancel those ions that appear on both sides of the equation. What is left will be the NET ionic equation.
Let me know what you come up with.
6NH4Cl is soluble and it is written as 6NH4^+ + 6Cl^-
To be consistent, I should have written it as 6NH4^+(aq) + 6Cl^-(aq).
Ok the ionic equation I came up with is:
6NH4(aq)--->3Cu2(aq)
but if this equaton is the NET ionic equation then what is the complete ionic equation?Because I need to finf the spectator ions and I don't think I can withput a complete ionic equation.
Absolutely not.
Reread my last two posts. They tell you want to do.
You write 3CuCl2 as 3Cu^+2(aq) + 6Cl^-(aq)
You write 2(NH4)3PO4 as 6NH4^+(aq) + 2PO4^-3(aq).
You write Cu3(PO4)2 as Cu3(PO4)2(s).
You write 6NH4Cl as 6NH4^+(aq) + 6Cl^-(aq).
Write that down on a sheet of paper. Her is what it will look like but check it out to make sure.
3Cu+2(aq) + 6Cl^-(aq) + 6NH4^+(aq) + 2PO4^-3(aq) ==> Cu3(PO4)2(s) + 6NH4^+(aq) + 6Cl^-(aq)
Again, check it out but the above should be the ionic equation.
NOW, look at the ionic equation and cancel the ions that are the same on both sides of the equation.
Do you have 3Cu^+2(aq) on both sides of the equation? No. So leave that.
Do you have 6Cl^-(aq) on both sides of the equation. Yes. Cancel those.
Do you have 6NH4^+(aq) on both sides of the equation. Yes. Cancel them.
Do you have 2PO4^-3(aq) on both sides of the equation? No. Leave it.
Do you have Cu3(PO4)2(s) on both sides of the equation? No. Leave it.
What do you end up with for the Net ionic equation?
I'm sorry I'm having so much trouble!So let me get this:
The balanced molecular equation is:
3CuCl2+2(NH4)3PO4--->Cu3(PO4)2+6NH4Cl
Then the complete ionic equation is:
3Cu2(aq)+6Cl-(aq)+6NH4(aq)+2PO4-3(aq)--->Cu3(PO4)2(s)+6NH4(aq)+6Cl-(aq)
so the NET ionic equation is:
3Cu2(aq)+2PO4-3(aq)---> Cu3(PO4)2(s)
If all this is right now can you help me find the spectator ions in this reaction because this is the last thing I need to do.
Then the complete ionic equation is:
3Cu2(aq)+6Cl-(aq)+6NH4(aq)+2PO4-3(aq)--->Cu3(PO4)2(s)+6NH4(aq)+6Cl-(aq)
The equation is ok except you left the + charge off the 6NH4^+(aq) on the right side.
Everything else looks ok. The spectator ions are those you canceled. That is why they are called "spectator" ions--because they didn't enter into the reaction. Just like spectators at a football game. They see what is going on but they don't get involved in the ball game.
Thank you so much for helping me get through the whole problem!Sorry it took me so long to understand!
You're welcome. Come again.
Here are the step-by-step solutions to the problem:
1. Write the balanced molecular equation for this reaction:
CuCl2 + (NH4)3PO4 -> Cu3(PO4)2 + 6NH4Cl
2. Write the balanced complete ionic equation for this reaction:
3Cu^2+(aq) + 6Cl^-(aq) + 6NH4+(aq) + 2PO4^3-(aq) -> Cu3(PO4)2(s) + 6NH4+(aq) + 6Cl^-(aq)
3. Spectator ions are ions that appear on both sides of the equation and do not participate in the reaction. In this case, the spectator ions are:
6NH4+(aq) and 6Cl^-(aq)
4. Write the net ionic equation for this reaction:
3Cu^2+(aq) + 2PO4^3-(aq) -> Cu3(PO4)2(s)
I hope this helps! Let me know if you have any further questions.
1. The balanced molecular equation for this reaction is: CuCl2 + (NH4)3PO4 = Cu3(PO4)2 + NH4Cl. To balance this equation, start by making sure that the number of atoms on each side of the equation is equal. In this case, we have 3 copper atoms on the left side and 3 copper atoms on the right side. We also have 8 chlorine atoms on the left side and 8 chlorine atoms on the right side. Lastly, we have 12 hydrogen atoms on the left side and 12 hydrogen atoms on the right side.
2. The balanced complete ionic equation for this reaction is: 3Cu^2+(aq) + 6Cl^-(aq) + 6NH4^+(aq) + 2PO4^3-(aq) = Cu3(PO4)2(s) + 6NH4^+(aq) + 6Cl^-(aq). To write the complete ionic equation, start by separating all of the soluble compounds into their ions. In this case, CuCl2 dissociates into 3Cu^2+(aq) and 6Cl^-(aq), and (NH4)3PO4 dissociates into 6NH4^+(aq) and 2PO4^3-(aq).
3. The spectator ions in this reaction are NH4^+(aq) and Cl^-(aq). Spectator ions are those ions that appear on both sides of the equation and do not participate in the actual reaction. In this case, NH4^+(aq) and Cl^-(aq) are present in both the reactants and products and do not undergo any changes.
4. The net ionic equation for this reaction is obtained by canceling out the spectator ions. Since NH4^+(aq) and Cl^-(aq) are the spectator ions, they are canceled out on both sides of the equation. The net ionic equation is: 3Cu^2+(aq) + 2PO4^3-(aq) = Cu3(PO4)2(s). This equation represents only the ions that are directly involved in the reaction.