1. You add 500 ml of 0.100 M AgNO3 solution to a solution containing an excess of Cl- ion. How much
AgCl precipitate will you form?
2. If you mix 200 ml of 0.100 M Pb(NO3)2 and 300 ml of 0.200 M MgCl2, how much PbCl2 precipitate
will you form?
3. A 0.150 g sample of solid lead(II) nitrate is added to 125 mL of 0.100 M sodium iodide solution.
Assume no change in volume of the solution. The chemical reaction that takes place is represented by
the following equation.
(a)
(b)
(c)
(d)
(e)
Pb(NO3)2(s) + 2 NaI(aq)  PbI2(s) + 2 NaNO3(aq)
List an appropriate observation that provides evidence of a chemical reaction between
the two compounds.
Calculate the number of moles of each reactant.
Identify the limiting reactant. Show calculations to support your identification.
Calculate the molar concentration of NO3–(aq) in the mixture after the reaction is
complete.
Circle the diagram below that best represents the results after the mixture reacts as
completely as possible. Explain the reasoning used in making your choice.
4. Answer the following questions about acetylsalicylic acid, the active ingredient in aspirin.
(a) The amount of acetylsalicylic acid in a single aspirin tablet is 325 mg, yet the tablet has a
mass of 2.00 g. Calculate the mass percent of acetylsalicylic acid in the tablet.
(b) The elements contained in acetylsalicylic acid are hydrogen, carbon, and oxygen. The
combustion of 3.000 g of the pure compound yields 1.200 g of water and 3.72 L of dry
carbon dioxide, measured at 750. mm Hg and 25C. Calculate the mass, in g, of each
element in the 3.000 g sample.
(c) A student dissolved 1.625 g of pure acetylsalicylic acid in distilled water and titrated the
resulting solution to the equivalence point using 88.43 mL of 0.102 M NaOH(aq). Assuming
that acetylsalicylic acid has only one ionizable hydrogen, calculate the molar mass of the
acid.
5. What mass (g) of CaF2 is formed when 47.8 mL of 0.334 M NaF is treated with an excess of aqueous
calcium nitrate?
6. What volume (mL) of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl?
7. What are the respective concentrations (M) of Fe3+ and I- afforded by dissolving 0.200 mol FeI3 in
water and diluting to 725 mL?
ANSWERS
1. 0.0500 moles AgCl
2. 0.0200 moles PbCl2
3. (a) formation of a yellow ppt
1 mol
(b) 0.150 g Pb(NO3)2 
= 4.5310-4 mol Pb(NO3)2
331.2 g
0.100 mol
0.125 mL 
= 0.0125 mol NaI
1L
(c) Pb(NO3)2; since they react in a 1:2 ratio of Pb(NO3)2:NaI, 6.2510-3 mol Pb(NO3)2 would
be required to react with all the NaI, therefore, you would run out of the lead nitrate first
2 mol NO-3
4.53  10 3 mol Pb(NO 3 )2 
1 mol Pb(NO 3 )2
(d) the nitrate is a spectator ion,
= 7.2510-3 M
0.125 L
(e)
the solution contains the sodium and nitrate spectator ions as well as
some excess iodide.
0.325 g
2.00 g  100% = 16.3%
(1.0079)(2) g H
(b) 1.200 g H2O  (1.0079)(2) + 16 g H2O) = 0.134 g H
750
760 atm (3.72 L)
P•V
n = R•T = (0.0821L•atm·mol-1•K-1)(298 K) = 0.150 mol CO2
12.0 g C
0.150 mol CO2  1 mol CO = 1.801 g C
2
3.000 g ASA – (1.801 g C + 0.134 g H) = 1.065 g O
0.102 mol
(c) 0.08843 L 
= 0.00902 mol base
1L
1 mol base = 1 mol acid
1.625 g ASA
0.00902 mol = 180 g/mol
4. (a)
5. 0.623 g
6. 13.1 mL
7. 0.276 M and 0.828 M