# 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+

moles AgNO3 = 0.02 mol AgNO3

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 AgCl

The moles of AgCl formed by CaCl2

= 0.03 mol AgCl

AgNO3 - limiting reactant - forms less AgCl

The mass of AgCl formed:

= 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

99% (247 ratings) ###### 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.