Problem: What mass of silver chloride can be prepared by the reaction of 100.0 mL of 0.20 M silver nitrate with 100.0 mL of 0.15 M calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.

FREE Expert Solution

Balanced equation: 2AgNO3 + CaCl2  2AgCl + Ca(NO3)2

  • AgCl → insoluble → precipitate
  • All Nitrates are soluble
  • salts of Cl-  are soluble except when paired with Ag+, Pb2+, Cu+


Molarity (M) =moles of soluteLiters of solution



moles AgNO3=100.0 mL×10-3 L1 mL×0.20 mol AgNO3L

moles AgNO3 = 0.02 mol AgNO3


moles CaCl2=100.0 mL×10-3 L1 mL×0.15 mol CaCl21 L

moles CaCl= 0.015 mol CaCl2


limiting reactant - reactant that forms the less amount of product

  • determines the maximum moles of product (AgCl(s))


mole-to-mole comparison:

  • 2 moles AgNOform 1 mole AgCl
  • 1 mole CaCl2 forms 1 mole AgCl


The moles of AgCl formed by AgNO3


0.02  mol AgNO3×2 mol AgCl2 mol AgNO3= 0.02 mol AgCl


The moles of AgCl formed by CaCl2


0.015  mol CaCl2×2 mol AgCl1 mol CaCl2= 0.03 mol AgCl


AgNO3 - limiting reactant - forms less AgCl


The mass of AgCl formed:


0.02  mol AgCl×143.32 g AgCl1 mol AgCl= 2.87 g AgCl


The mass of silver chloride that can be prepared by the reaction is 2.87 g.


Ions remaining in solution after precipitate forms:

  • CaCl→ Ca2+ + 2Cl-
  • Ca(NO3)2 → Ca2+ + 2NO3-
    • AgNO3 is used up in the reaction


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Problem Details

What mass of silver chloride can be prepared by the reaction of 100.0 mL of 0.20 M silver nitrate with 100.0 mL of 0.15 M calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.

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