Question

13.A student mixes 100.0 mL of 0.500 M AgNO₃ with 100.0 mL of 0.500 M CaCl₂. (8 points)

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?

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

Answer 1

a)

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

b)

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

c)

calculate moles of Ag+ and moles of Cl-

moles of Ag = V*M = 0.1*0.5 = 0.05 mol of Ag+

moles of CaCl2 = 2*V*M = 2*0.1*0.5 = 0.10 mol of Cl-

limiting reactant is Ag+, expect only 0.05 mol of AgCl precipitat

MW AgCl = 143.32

mass = mol*MW = 0.05*143.32 = 7.16 g

d)

Final concentrations

Vt = V1+V2 = 100 ml + 100 ml = 200 ml of 0.2

Find moles:

Ag = 0 (all reacted)

NO3 = 0.05 mol

Ca = 0.05

Cl2 = 0.10 - 0.05 = 0.05

Calculate concentration (divide moles/Vt which is 0.2)

[Ag-] = 0 (all reacted)

[NO3-] = 0.25

[Ca+2] = 0.25

[Cl-] = 0.25