ead (II) ions can be removed from solution with KCl according to the reaction: Pb2+ (aq) + KCl (aq) ⟶ PbCl2 (s) + K+ (aq) When 28.5 g KCl is added to a solution containing 25.7 g Pb2+, a PbCl2 (s) forms. The solid is filtered and dried and found to have a mass of 29.4 g. Determine the limiting reactant, the theoretical yield of PbCl2, and the percent yield for the reaction.
ead (II) ions can be removed from solution with KCl according to the reaction:
Pb2+ (aq) + KCl (aq) ⟶ PbCl2 (s) + K+ (aq)
When 28.5 g KCl is added to a solution containing 25.7 g Pb2+, a PbCl2 (s) forms. The solid is filtered and dried and found to have a mass of 29.4 g. Determine the limiting reactant, the theoretical yield of PbCl2, and the percent yield for the reaction.
Firat we would balance the given chemical reaction.
Then to calculate limiting reagent, we would calculate moles of PbCl2 from both reactants using their mass and using mole ratio. The reactant which will produce less moles of PbCl2 will be limiting reagent.
Theoretical yield would be calculated using the moles of PbCl2 produced from limiting reagent. Molar mass would be used to calculate mass from moles.
Percent yield can be calculated using actual yield and theoretical yield.
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