Pb(NO3)2 + 2KCl -------------> PbCl2 + 2KNO3
331.2 g Pb(NO3)2 reacts with 2 x 74.55 g KCl
5.22 g Pb(NO3)2 reacts with 5.22 x 2 x 74.55 / 331.2 = 2.35 g KCl
but we have 5.19 g KCl so KCl is exess reagent and Pb(NO3)2 is limiting reagent.
limiting reagent = lead (II) nitrate
331.2 g Pb(NO3)2 forms 271.8 g PbCl2
5.22 g Pb(NO3)2 forms 5.22 x 271.8 / 331.2 = 4.28 g
% yield = 90.3
90.3 = (X / 4.28) x 100
X = 3.86 g
mass of precipitate = 3.86 g
mass of exess reagent = 5.19 - 2.35 = 2.84 g
exess reagent left = 2.84 g
An aqueous solution containing 5.22 g of lead(II) nitrate is added to an aqueous solution containing...
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