Solving Limiting Reactant problems in solution

Question

Suppose  0.0187g of sodium iodide is dissolved in 50.mL of a 22.0mM aqueous solution of silver nitrate.Calculate the final molarity of sodium cation in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. Round your answer to 2 significant digits.

Answer 1

First, write the correctly balanced equation:

NaI(s) + AgNO₃(aq) ==> NaNO₃(aq) + AgI(s)

Next, find the limiting reactant. The easiest way to do this (there are several ways) is to divided the moles of each reactant by the coefficient in the equation, and whichever is less is limiting. Thus...

For NaI we have 0.0187 g x 1 mol NaI/149.89 g = 0.00012476 moles NaI (÷ 1 = 0.00012476)

For AgNO3 we have 50.0 ml x 1 L/1000 ml x 0.022 mol/L = 0.0011 moles AgNO3 (÷ 1 = 0.0011)

So, we conclude that NaI is limiting.

We are asked to find the final [Na+]. The Na+ does NOT end up in the precipitate of AgI, so all of the Na+ from the NaI will be in NaNO3 and will be present as free Na+. Since we already found there to be

0.00012476 moles NaI, there will be 0.00012476 moles Na+ as well.

Final [Na+] = 0.00012476 moles Na+ / 0.050 L = 0.002495 mol/L = 0.0025 M (to 2 significant figures)

You can also report this as 2.5 mM