Hi Margie!
Well, let us work with each question: (Using Double-Substitution Method)
(a). CuSO4 (aq) + Na2CO3 (aq) --> CuCO3 (aq) + Na2SO4 (aq)
When both the reactants are mixed, carbonate is soluble with Cu (its exceptions are Group 1 elements and NH4+), and sulfate is soluble with Na (its exceptions are Ag+, Ca2+, Sr2+, Ba2+, Pb2+). So no precip will occur.
(b). HI (aq) + Zn(NO3)2 (aq) --> ZnI2 (aq) + HNO3 (aq)
When both the reactants are mixed, iodide [halides] is soluble with Zn (its exceptions are Ag+, Pb2+, Hg(2)2+), and nitrate is soluble with H (there are no exceptions). So no precip will occur.
So far, both (a) and (b) have yielded no precip reactions. However, (c) is exciting:
(c). AgNO3 (aq) + NaBr (aq) --> NaNO3 (aq) + AgBr (s)
So far, when both the reactants are mixed, nitrate can be soluble with Na, but for Br, the exceptions for halides are there following:
Halides can be soluble in anything except Ag+, Pb2+, Hg(2)2+.
So that would conclude that reaction c will produce some form of precipitation coming from the AgBr (s).
Now the completed balanced reaction and net ionic equation are below.
Balanced Equation:
AgNO3 (aq) + NaBr (aq) --> NaNO3 (aq) + AgBr (s)
The equation is already balanced because you can count equal terms for each compound.
Net Ionic Equation:
We need to decompose the Balanced equation above by working with ions in the aqueous solution.
*AgBr (s) will not decompose because it is solid, not aqueous.*
**We will use the definition of spectator ions - the ions that appear the same on both sides of the equation.**
Ag+1 (aq) + NO3-1 (aq) + Na+1 (aq) + Br-1 (aq) --> Na+1 (aq) + NO3-1 (aq) + AgBr (s)
Now we see that using the definition of spectator ions; we can eliminate like terms on both sides.
Ag+1 (aq) + NO3-1 (aq) + Na+1 (aq) + Br-1 (aq) --> Na+1 (aq) + NO3-1 (aq) + AgBr (s)
And this is your Net Ionic Equation:
Ag+1 (aq) + Br-1 (aq) --> AgBr (s)
Hopefully, this helps, and feel free to ask more questions/concerns, and another tutor or I will happily respond!
