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Write the complete and net ionic equations for the following reactions. Be sure to indicate the states of the reaction products.

1) AgNO3(aq) + NaCl(aq) ---->

2) CaCl2(aq) + Na2SO4(aq) ---->

3) Pb(ClO4)2(aq) + NaOH(ag) ---->

4) (NH4)2SO4(aq) + BaCl2(aq) ---->

2 answers

I'll do #1.
Step 1. Write the molecular equation.
AgNO3(aq) + NaCl(aq) = AgCl + NaNO3

Step 2. If a reaction occurs, it is because
a. a precipitate forms (an insoluble material forms). For this you need to know the solubilities of the salts. Here is a simplified chart for solubility.
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
b. A gas is formed.
c. A slightly ionized material is formed (usually water).

Step 3. Now you can take the products and add their state.
AgNO3(aq) + NaCl(aq) ==> AgCl(s) + NaNO3(aq)
You will note from the solubility chart that AgCl is not soluble and NaNO3 is; therefore, I place a (s) by AgCl and (aq) by NaNO3.

Step 4. Next you convert to the FULL ionic equation. (Anything NOT in step 2a,b, or c gets converted to ions.)
Ag^+ + NO3^- + Na^+ + Cl^- ==> AgCl(s) + Na^+ + NO3^-
Note: All of the soluble materials should have (aq) by them but that clutters the screen and I have omitted them. You need to write these in for your instructor. For example, part of the equation would look like this
Ag^+(aq) + NO3^-(aq) + Na^+(aq) + Cl^-(aq) etc etc. AgCl does NOT have (aq) since it is insoluble but the NaNO3 does.

Step 5. Cancel ions common to both sides. I see Na^+ on both sides and I see NO3^- on both sides. Cancel those.

Step 6. What is left is the NET ionic equation.
Ag^+(aq) + Cl^-(aq) ==> AgCl(s)
Students often ask what is the purpose of a net ionic equation. This last equation shows you what actually happens in the solution. The Ag^+(aq) reacts with Cl^(aq) to form insoluble AgCl(s). What about the other ions? It doesn't matter what they are. They are just spectator ions. They don't enter into the reaction. So AgNO3 + NaCl gives the same net ionic equation as AgNO3 and KCl or LiCl or any other soluble chloride.
THANK YOU!!!!