Question

A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution, Pb(NO3)2 (aq) is added to precipitate out 0.7822 grams of PbBr2 (s). What is the percent by mass of potassium bromide in the mixture? What is the percent by mass of strontium nitrate in the mixture?

Answer 1

2KBr(aq) + Pb(NO3)2(aq) ----------> PbBr2(s) +2KNO3(aq)

no of moles of PbBr2   = W/G.M.Wt

                                      = 0.7822/367   = 0.00213moles

1 mole fo PbBr2 produced from 2 moles of KBr

0.00213 moles of PbBr2 produced from = 2*0.00213/1   = 0.00426 moles of KBr

mass of KBr = no of moles * gram molar mass

                         = 0.00426*119 = 0.507g

percent by massof KBr in the mixture = 0.507*100/5   = 10.14%

percent by mass of Sr(NO3)2   = 100-10.14   = 89.86% Sr(NO3)2