Problem: Lead ions can be precipitated from solution with KCl according to the following reaction: Pb2+ (aq) + 2 KCl (aq) → PbCl2 (s) + 2 K+ (aq)When 28.6 g KCl is added to a solution containing 25.7 g Pb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.4.   Determine the limiting reactant.

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Balanced equation: Pb2+ (aq) + 2 KCl (aq) → PbCl2 (s) + 2 K+ (aq)


Mole to mole comparison

  • 1 mole Pb2+ forms 1 mole PbCl2
  • 2 moles KCl  form 1 mole PbCl2


limiting reactant: forms the less amount of product 


The mass of PbCl2 formed by KCl is: 

  • Molar Mass of KCl = 74.55 g/mol 
  • Molar Mass of PbCl278.11 g/mol 


28.6  g KCl×1 mol KCl74.55  g KCl×1 mol PbCl22 mol KCl×278.11 g PbCl21 mol PbCl2


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Problem Details

Lead ions can be precipitated from solution with KCl according to the following reaction: 

Pb2+ (aq) + 2 KCl (aq) → PbCl2 (s) + 2 K+ (aq)

When 28.6 g KCl is added to a solution containing 25.7 g Pb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.4.   

Determine the limiting reactant.

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