Question

The reaction of aqueous potassium chloride and aqueous lead(II) nitrate produces lead(II) chloride and potassium nitrate, according to the balanced chemical equation shown. 2 KCl (aq) + Pb(NO3)2 (aq) → PbCl2 (s) + 2 KNO3 (aq) If equal volumes of 0.500 M KCl (aq) and 0.500 M Pb(NO3)2 (aq) are combined, what will the final concentration of KNO3 (aq) be?

Answer 1

2 KCl (aq) + Pb(NO3)2 (aq) ----> PbCl2 (s) + 2 KNO3 (aq)

let us assume the volume of KCl = 20 mL = Pb(NO3)2

number of moles of KCl = 20 * 0.5 = 10 mmol

number of moles of Pb(NO3)2 = 20 * 0.5 = 10 mmol

2 mole of KCl Produces 2 mole of KNO3

10 mmol of KCl produces 2/2*10 mmol of KNO3

                         = 10 mmol of KNO3

1 moles of Pb(NO3)2 produces 2 moles of KNO3

10 mmol of Pb(NO3)2 produces 2/1*10 mmol of KNO3

                                               = 20 mmol of KNO3

so KCl is limting reagent.

final concentration of KNO3 = (10 * 10^-3*1000/(20+20))

                             = 0.25 mol/L