posted by Drake on .
Help writing complete ionic and net ionic equations.?
Hg2(NO3)2 (aq) + CuSO4 ---> Cu(NO3)2 (aq) + Hg2SO4 (s)
K2CO3 (aq) + MgI2 --> MgCO3 (s) + 2KI
3 NaCrO4 (aq) +2 AlBr3 ---> Al2(CrO4)3 (s) + 6NaBr
i got down writing the formula equation but i would like help with the complete and net ionic equations i just don't get it and would like help
Here is what you do. First write and balance the molecular equation. I think you have done that.
K2CO3(aq) + MgI2(aq) --> MgCO3(s) + 2KI(aq)
Step 2. Convert those aq solns to ions. I will omit the aq part on each ion t save typing but you should put it in. If it isn't aq (for examle the MgCO3 is (s), you leave it as is. Slightly ionized materials such as H2O(l) would not be changed.) This gives you the COMPLETE ionic equation.
2K^+ + CO3^2- + Mg^2+ + 2I^- ==> MgCO3(s) + 2K^+ + 2I^-
Step 3. Cancel ions common to each side. I see 2K^+ on both sides and 2I^- on both sides. Cancel those. Those are the spectator ions. What is left is the NET ionic equation.
Mg^2+(aq) + CO3^2-(aq) ==> MgCO3(s)
Three steps will do them all.
Write and balance the molecular equation.
Separate the (aq) materials (those soluble in water) into ions. This give you the complete ionic equation.
Last step is to cancel the ions common to both sides. What is left is the net ionic equation.