Question

A solution is prepared by mixing 45.00 mL of 0.022 M AgNO3 with 13.00 mL of 0.0014 M Na2CO3. Assume volumes are additive.
Calculate [Ag+], [CO32-], [Na+] and [NO3-] after equilibrium is established.

Answer 1

....2AgNO3 + Na2CO3 ==> Ag2CO3 + 2NaNO3
mmols AgNO3 = 45.00*0.022 = 0.99
mmols Na2CO3 = 13.00*0.0014 = 0.0182.

As a limiting reagent (LR) problem we could form 0.0182 mol Ag2CO3 with Na2CO3 or 0.495 mol Ag2CO3 with AgNO3. That makes Na2CO3 the LR. So we use up all of the Na2CO3 (at least all of the CO3^2-) and 0.0182*2 = 0.0364 mol AgNO3 which leaves 0.99-0.364 = 0.954 mols AgNO3 in excess.
I assume you are not to worry about Ksp of Ag2CO3(s) since the Ag^+ from AgNO3 is a common ion and will affect solubility of Ag2CO3.
Total volume is 58 mL. We are assuming all of the carbonate is gone.
(Ag^+) = 0.954/58 = ?
etc.