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Lead ions can be precipitated from solution with KCl according to the following reaction:
Lead ions can be precipitated from solution with KCl according to the following reaction:Pb2+(aq) + 2KCl(aq) --> PbCl2(s) + 2K+(aq)
When 28.8g KCl is added to a solution containing 25.6g Pb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.6g.
Determine the theoretical yield of PbCl2.
m= ____________g
Determine the percent yield for the reaction.
Answer:34.48 grams = Theoretical yieldThe % yield is 85.85 %Explanation:Step 1:The balanced equationPb2+(aq) + 2KCl(aq) → PbCl2(s) + 2K+(aq)For 1 mole Pb2+ consumed, we need 2 moles of KCl to produce 1 mole of PbCl2 and 2 moles of K+Step 2:Given dataMass of KCl = 28.8 gramsMass of Pb2 = 25.6 gramsMass of dried PbCl2 = 29.6 gramsMolar mass of Pb2+ = 207.2 g/molMolar mass of KCl = 74.55 g/molMolar mass of PbCl2 = 278.1 g/molStep 3:Calculate moles of Pb2+Moles = mass / Molar massMoles Pb2+ = 25.6 grams / 207.2 g/mol = 0.124 molesStep 4:Calculate moles of KClMoles = 28.8 grams / 74.55 g/mol = 0.386 molesStep 5: Calculate limiting reactantFor 1 mole Pb2+ consumed, we need 2 moles of KClPb2+ is the limiting reactant. It will completely ( 0.124 moles) be consumed. There will remain 0 molesThere will react 2* 0.124 = 0.248 moles of KCl. There will remain 0.386 - 0.248 = 0.138 moles of KClStep 6:Calculate number of moles of PbCl2For 1 mole Pb2+ consumed, we need 2 moles of KCl to produce 1 mole of PbCl2 and 2 moles of K+For 0.124 moles of Pb2+ consumed ,there will be produced 0.124 moles of PbCl2Step 7:Calculate mass of PbCl2Mass = Numbers of moles * Molar massmass PbCl2 =0.124 moles * 278.1 g/mol = 34.48 grams = Theoretical yieldStep 8: Calculate % yield% yield = (actual yield / theoretical yield) * 100 = (29.6 grams / 34.48 grams) *100 % = 85.85 %The % yield is 85.85 %...