Question

13.A student mixes 100.0 mL of 0.500 M AgNO₃ with 100.0 mL of 0.500 M CaCl₂.

a.Write the balanced molecular equation for the reaction.

b.Write the net ionic equation for the reaction.

c.How many grams of precipitate will form?

d.What is the concentration of Ag⁺, NO₃^‒, Ca²⁺, and Cl^‒ in the final solution (assume volumes are additive).

Answer 1

moles of AgNO3 = 100 x 0.5 / 1000 = 0.05

moles of CaCl2 = 100 x 0.5 / 1000 = 0.05

a) balanced molecular eqaution :

2 AgNO3 + CaCl2 ----------------> 2 AgCl + Ca(NO3)2

b) net ionic equation :

2 Ag+ + 2 Cl- ----------------> 2 AgCl (s)

c)

2 AgNO3 + CaCl2 ----------------> 2 AgCl + Ca(NO3)2

2 1 2 1

0.05 0.05 ??

here limiting reagent is AgNO3. so product formed according to that.

2 moles AgNO3 ----------------> 2 moles AgCl precipitate

0.05 moles AgNO3 -------------- > ??

moles of AgCl precipitate = 0.05 moles

moles = mass / molar mass

0.05 = mass / 143.3

mass of precipitate = 7.165 g

d)

2 AgNO3 + CaCl2 ----------------> 2 AgCl + Ca(NO3)2

2 1 2 1

0.05 0.05 ??

moles of Ca(NO3)2 produced = 0.025 moles .

moles of CaCl2 remained = 0.025 moles

[Ca+2] = moles of Ca in Ca(NO3)2 + moles of Ca in CaCl2 / total volume

= 0.025 + 0.025 / 0.2

= 0.25 M

[NO3-] = moles of NO3- / total volume

= 2 x 0.025 / 0.2

= 0.25 M