Balanced equation: 2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2
moles AgNO3 = 0.02 mol AgNO3
moles CaCl2 = 0.015 mol CaCl2
limiting reactant - reactant that forms the less amount of product
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:
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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