Question

Suppose 39.4g of zinc chloride is dissolved in 250.mL of a 0.60 M aqueous solution of potassium carbonate. Calculate the final molarity of zinc cation in the solution. You can assume the volume of the solution doesn't change when the zinc chloride is dissolved in it. Round your answer to 2 significant digits.

Answer 1

no of moles of ZnCl2   = W/G.M.Wt

                                = 39.4/136.286   = 0.289moles

no of moles of K2CO3 = molarity * volume in L

                                  = 0.6*0.25    = 0.15moles

ZnCl2(aq) + K2CO3(aq) ---------> ZnCO3(s) + 2KCl(aq)

1 moles of K2CO3 react with 1 mole of ZnCl2

0.15moles of K2CO3 react with 0.15 moles of ZnCl2 is required

ZnCl2 is excess reactant

The no of moles of excess reactant left after complete the reaction = 0.289-0.15   = 0.139moles

molarity of ZnCl2   = no of moles/volume in L

                          = 0.139/0.25   = 0.56M

ZnCl2(aq) -----------> Zn^2+(aq) + 2Cl^- (aq)

0.56M ------------------ 0.56M

The final molarity of Zn^2+ = 0.56M >>>>answer