What is the net ionic equation for the reaction copper(II) chloride(aq) + ammonium carbonate(aq)?

Question

matt101 · Answer

The net ionic equation includes only things that CHANGE their phase during the reaction (usually going from aqueous to solid). The ions that don't change their phase and remain dissolved are called spectator ions. The ionic equation includes the spectator ions in addition to the ions that undergo a phase change.

Before we worry about the net ionic equation, we need to come up with the full reaction. What is the balanced chemical reaction for the chemicals provided in the question?

anonymous · Answer

Wouldn't that be 3(NH4)2CO3 + 2CuCl3 = 6NH4Cl + Cu2(CO3)3?

matt101 · Answer

It said copper (II) chloride so it would be

(NH4)2CO3 (aq) + CuCl2 (aq) = 2 NH4Cl (aq) + CuCO3 (s)

Makes the equation a bit simpler doesn't it! I also added in the phases for all the compounds in the equation. The aq indicates the compound is aqueous (dissolved), and so it has dissociated into its component ions. The s indicates the compound is a solid, so the compound precipitates out of solution and is no longer dissociated. What would be the IONIC equation for this reaction?

anonymous · Answer

Would it be something like this? NH4^+ + CO3^-2 + Cu^+2 + Cl2^- = NH4^+ + Cl^- + Cu^+2 +CO3^-2?

matt101 · Answer

Very close! Two things - the equation needs to stay balanced, so you need to have 2 NH4+ and 2 Cl- on the left and right sides. And keep in mind that CuCO3 has PRECIPITATED. In other words, it is no longer dissolved and split into its ions. So your ionic equation would really be:

2 NH4^+ + CO3^-2 + Cu^+2 + 2 Cl^- = 2 NH4^+ + 2 Cl^- + CuCO3?

The net ionic equation takes out the spectator ions - the ones that don't change from the left to the right side. When you take out the guys that don't change their phase, what are you left with?

anonymous · Answer

So that would be CO3^-2 + Cu^+2 = CuCO3!

matt101 · Answer

Yup :)

matt101 · Answer

It's good to include the states as well in these equation (e.g. aq, s, etc)

anonymous · Answer

Ok thank you so much for your help, I finally understand the process!