Question

A 250 mL solution of 2.5 M Potassium Hydroxide is mixed with a 50 mL solution of 3.0 M Barium Chloride and 5.0 M Calcium Bromide. The mixture of these solutions results in the formation of a precipitate. What is the precipitate and how much of it forms, assuming that the insoluble product precipitates completely?

Answer 1

balanced equation :      CaBr2(aq) + 2KOH(aq) ---> Ca(OH)2 (s) + 2KBr(aq)

in the mixture there is two products posiible.such as Ba(OH)2 -soluble

              Ca(OH)2 - insoluble - precipitate
  

1 mol CaBr2(aq) = 2 mol KOH(aq)

no of mol of KOH(aq) reacted = 250*2.5/1000 = 0.625 ml

no of mol of CaBr2(aq) reacted =50*5/1000 = 0.25 mol

no of mol of Ca(OH)2 (s) precipitate = 0.25 mol

mass of Ca(OH)2 (s) precipitate = 0.25 *74 = 18.5 g