Question

You have 75.0mL of a 2.50 M solution of Na2CrO4 (aq). You also have 125mL of a 1.88 M solution of AgNO3 (aq). Calculate the concentration of NO3- after the two solutions are mixed together

Answer 1

First, write out the balanced equation for the reaction: 
Na2CrO4 + 2AgNO3 --> Ag2CrO4 + 2NaNO3 

Second, convert the given concentrations into moles by converting the volume into liters and multiplying it by the molarities. 1000mL = 1L
(.075L)(2.50M) = 0.1875 mol Na2CrO4
(.125L)(1.88M) = 0.235 mol AgNO3

Third, figure out the limiting reactant in the reaction. In this case, the limiting reactant is AgNO3

Fourth, using the mole ratios and stoichiometry, convert moles of AgNO3 into moles of NaNO3:
(0.235 mol AgNO3) (2mol NaNO3/2 mol AgNO3) = 0.235 mol NaNO3

Now to find the concentration of the NO3 ion, we can assume that the volume is now .200L (.075L + .125L) since the solutions are being mixed together.
Molarity = mol/L, so (0.235 mol NaNO3)/(.200L) = 1.18M NO3- ion