Ionic equations

But the 3rd equation is the result of cancelling out the spectator ions.

OK, so first let's write out the equation:

1) Na₂CO₃ + 2HNO₃ ------> 2NaNO₃ + H₂O + CO₂

Now let's separate everything into its constituent ions:

2) 2Na⁺ + CO₃^2- + 2H⁺ + 2NO₃^- ------> 2Na⁺ + 2NO₃^- + H₂O + CO₂

Now, ions or species which are on both sides of the arrow can be cancelled therefore final answer is:

3) CO₃^2- + 2H⁺ ----> H₂O + CO₂.




Try this method with the HCl example and quote me if you're having problems

One I'm finding quite difficult is the reaction between Sodium sulfate and barium chloride:
Ba2+ + 2Cl- + 2Na+ + SO4-2 -------> Ba2+ + 2SO4-2 + 2Na+ + 2Cl-

All ions are present on both sides, how do I know which are spectators and which aren't?

I'm going to try and help based on what I did in Chemistry A Level a couple of years ago.

On the right you've just added in an extra SO₄^2- out of nowhere.

If I remember:
BaCl₂ + Na₂SO₄ --> BaSO₄ + 2NaCl

So the ionic equation, including spectators, would be:

Ba²⁺ + 2Cl^- + 2Na⁺ + SO₄^2- -------> BaSO₄ + 2Na⁺ + 2Cl^-

This is the same equation as the one you wrote; except that (a) the extra SO₄^2- has been removed, and (b) I've not ionised the BaSO₄​, as it is a solid. Now I'll let you remove the spectators and come up with an ionic equation.

Thanks, the extra sulfate ion was a typo. The reason I couldn't do this was because BaSO4 is insoluble. Will there always be at least 1 compound which is either insoluble or non ionic?

The ionic equation should be:
Ba(2+)+SO4(2-)--->BaSO4(s)

I can't remember whether this is the case or not, but it may help to write the states of each ion or compound, to help.

Though chances are there would be - otherwise all of the ions would be the same each side of the arrow.

Hope I've helped!

Edit: Just checked and you have got the correct equation.

Ionic equation: CuO+2H(+)+2Cl(-)----->Cu(2+)+2Cl(-)+H2O(l)
Final equation: CuO(s)+2H(+)---->Cu(2+)+H2O(l)

Is this correct or completely wrong?

One I'm finding quite difficult is the reaction between Sodium sulfate and barium chloride:
Ba2+ + 2Cl- + 2Na+ + SO4-2 -------> Ba2+ + 2SO4-2 + 2Na+ + 2Cl-

All ions are present on both sides, how do I know which are spectators and which aren't?

In such cases think about solubility rules. Sometimes there is no reaction at all.

Ok, for the HCL one:

balanced equation: CuO(s)+2HCl(aq)---->CuCl2(aq)+H2O(l)
Is CuO solid state symbol since it's not soluble. Also, am I supposed to know it's insoluble and it won't dissociate into ions? Is there a rule for knowing when they do or don't when writing out ionic equations? Since CuCl is soluble, am I right in breaking that into ions?

Ionic equation: CuO+2H(+)+2Cl(-)----->Cu(2+)+2Cl(-)+H2O(l)
Final equation: CuO(s)+2H(+)---->Cu(2+)+H2O(l)

Is this correct or completely wrong?

Just replied earlier in that the BaSo₄ is a solid and is thus insoluble.

Yes, but checking solubility rules gives a general method, that can be applied to any reaction.

I do remember that there were solubility rules but can't remember the specifics as it was about 2 to 3 years ago that I did my A Level Chemistry.