CHAPTER 4 TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY
AP Chemistry released exam 1994
SECTION I
16. Commercial vinegar was titrated with NaOH solution to determine the content of acetic
acid, HC2H3O2. For 20.0 milliliters of the vinegar, 26.7 milliliters of 0.600-molar NaOH
solution was required. What was the concentration of acetic acid in the vinegar if no other
acid was present?
(A) 1.60 M
(B) 0.800 M
(C) 0.600 M
(D) 0.450 M
(E) 0.200 M
18. 2 H2O + 4 MnO4^- + 3 ClO2^- ->> 4 MnO2 + 3 ClO4^- + 4 OH^Which species acts as an oxidizing agent in the reaction represented above?
(A) H2O
(B) ClO4(C) ClO2(D) MnO2
(E) MnO422. HSO4^- + H2O <<->> H3O^+ + SO4^2In the equilibrium represented above, the species that act as bases include which of the
following?
I. HSO4^II. H2O
III. SO4^2(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III
50. Which of the following acids can be oxidized to form a stronger acid?
(A) H3PO4
(B) HNO3
(C) H2CO3
(D) H3BO3
(E) H2SO3
52. When dilute nitric acid was added to a solution of one of the following chemicals, a gas was
evolved. This gas turned a drop of limewater, Ca(OH)2, cloudy, due to the formation of a
white precipitate. The chemical was
(A) household ammonia, NH3
(B) baking soda, NaHCO3
(C) table salt, NaCl
(D) epsom salts, MgSO4 7H2O
(E) bleach, 5% NaOCl
53. If 87 grams of K2SO4 (molar mass 174 grams) is dissolved in enough water to make 250
milliliters of solution, what are the concentrations of the potassium and the sulfate ions?
[K^+]
[SO4^2-]
(A)
0.020 M
0.020 M
(B)
1.0 M
2.0 M
(C)
2.0 M
1.0 M
(D)
2.0 M
2.0 M
(E)
4.0 M
2.0 M
55. What volume of 0.150-molar Hcl is required to neutralize 25.0 milliliters of 0.120-molar
Ba(OH)2?
(A) 20.0 mL
(B) 30.0 mL
(C) 40.0 mL
(D) 60.0 mL
(E) 80.0 mL
56. It is suggested that SO2 (molar mass 64 grams), which contributes to acid rain, could be
removed from a stream of waste gases by bubbling the gases through 0.25-molar KOH,
thereby producing K2SO3. What is the maximum mass of SO2 that could be removed by
1,000. liters of the KOH solution?
(A) 4.0 kg
(B) 8.0 kg
(C) 16 kg
(D) 20. kg
(E) 40. kg
AP Chemistry released exam 1999
SECTION I
33. A 1.0 L sample of an aqueous solution contains 0.10 mol of NaCl and 0.10 mol of CaCl2.
What is the minimum number of moles of AgNO3 that must be added to the solution in
order to precipitate all of the Cl^- as AgCl(s) (Assume that AgCl is incoluble.)
(A) 0.10 mol
(B) 0.20 mol
(C) 0.30 mol
(D) 0.40 mol
(E) 0.60 mol
38. A molecule or an ion is classified as a Lewis acid if it
(A) accepts a proton from water
(B) accepts a pair of electrons to form a bond
(C) donates a pair of electrons to form a bond
(D) donates a proton to water
(E) has resonance Lewis electron-dot structures
56. A yellow precipitate forms when 0.5 M NaI(aq) is added to a 0.5 M solution of which of the
following ions?
(A) Pb^2+(aq)
(B) Zn^2+(aq)
(C) CrO4^2-(aq)
(D) SO4^2-(aq)
(E) OH^-(aq)
59. A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the
molar concentration of OH^-(aq) in the resulting solution? (Assume that the volumes are
additive.)
(A) 0.10 M
(B) 0.19 M
(C) 0.28 M
(D) 0.40 M
(E) 0.55 M
69. What is the final concentration of barium ions, [Ba^2+], in solution when 100. mL of 0.10
M BaCl2(aq) is mixed with 100.mL of 0.050 M H2SO4(aq)?
(A) 0.00 M
(B) 0.012 M
(C) 0.025 M
(D) 0.075 M
(E) 0.10 M
73. The volume distilled water that should be added to 10.0 mL of 6.00 M HCl (aq) in order to
prepare a 0.500 M HCl (aq) solution is approximately
(A) 50.0 mL
(B) 60.0 mL
(C) 100. mL
(D) 110. mL
(E) 120. mL
AP Chemistry 2004 free response question
Annotated Answers:
(a) (i) Solubility rules: all salts containing NO3 is soluble Pb(NO3)2 is not solution Q
all salts containing Na is soluble  NaCl is not solution Q
all salts containing CO3 is insoluble  forms precipitate  K2CO3 is the solution Q
(ii) K2CO3 + AgNO3  KNO3 + Ag2CO3(s) (not balanced)
K2CO3 + BaCl2  KCl + BaCO3 (s) (not balanced)
Therefore, Ag2CO3 and BaCO3 are the two precipitates
(b) (i) K2CO3 + Pb(NO3)2  KNO3 + PbCO3(s) (not balanced)  forms precipitate  Pb(NO3)2
is the solution R
K2CO3 + NaCl  Na2CO3 + KCl (not balanced)  no precipitate  NaCl is the solution
S
(ii) PbCO3 (refer to b(i)
Annotated Answers:
(c) (i) Mix solution S (NaCl) with all equal amounts of both Solution X and Solution Y into two
different labeled beakers
(ii) one of the solution will form precipitate while the other will not (NaCl with AgNO3 will form
precipitate, while NaCl with BaCl2 will not form any)
(iii) NaCl + AgNO3  NaNO3 + AgCl(s) (not balanced)
NaCl + BaCl2  NaCl + BaCl2 (not balanced)
if solution X forms precipitate, solution X is AgNO3, if not, solution Y will be AgNO3, and solution
X will be BaCl2
AP Chemistry 2005 free response question: Form B
Annotated Answers:
(a) In order to prepare 50 mL of 5.0 M solution of H2SO4, the student should first calculate the
number of moles of H2SO4. Then divided by the 10.0 M to get 25.0 mL of 10.0 M H2SO4 needed.
Then describe the procedures in lab. Goggles, 100 mL graduated cylinder, distilled water, 50.0 mL
volumetric flask, pipet will be needed accordingly.
(b) (2.7 g Al)(1 mol Al / 27.0 g Al)(4 mol H2SO4 / 2 mol Al)(1L / 5.0 mol H2SO4) = 0.040 L
Annotated Answers:
(c) (i) % yield = (experimental / theoretical) x 100%
if the crystals are not dried, it means that the excess water will make the mass greater, which will
make the calculated percent yield to high  therefore, the calculated percent yield will be greater
(ii) cooling the reaction will create more precipitation  higher mass of crystals  higher percent
yield
(d) since the crystals of pure alum is a hydrate, during the heating, the water will evaporate, which
the mass will decrease since the loss of water molecules.
AP Chemistry 2006 free response question
Annotated Answers:
(a) (i) find the number of moles of C in CO2 and find the mass with molar mass.
(ii) times the % of N to the total mass of the original sample
(iii) total mass minus the mass of H, C and N to find the mass of O.
(iv) convert all the masses to moles, and divide the number of moles with the least value and round
up to the integers to find the empirical formula
(b) gas law not in CH 4
AP Chemistry 2008 free response question
(a) the mass of the sample did not change significantly between the 2nd and the 3rd heatings
(b) (i) first calculate the mass of H2O lost, and divide by the molar mass of H2O to find the number
of moles of H2O
(ii) first find the mass of anhydrous MgCl2 by minus the mass after the 3rd heating by the empty
container. Then find the moles of MgCl2 by dividing the molar mass from the mass of MgCl2. Find
the ratio between moles of H2O and the moles of MgCl2 to find the hydrated compound formula
Annotated Answers:
(c) the mass of water lost by the hydrate will calculated larger because those mass of solid lost
during heating will be considered as water
(d) filtering the mixture  drying the precipitate  determining the mass by difference (lab
procedures)
(e) (i) find the number of moles of AgCl and use the mole ratio 1:2 to calculate the original mole
number
(ii) use the original mole number to find the mass of MgCl2 and MgCl2 / total mass x 100% to find
AP Chemistry 2010 free response question: Form B
Annotated Answers:
Part (a), (b), (c), are not included in CH4
Annotated Answers:
(d) find the mole ratio between Fe and Cr2O7 (6:1) and find the number of moles of Cr2O7
(e) the total amount add – excess mol of Cr2O7 will give the number of mole which reacted with
TeO2. To find the total number of moles of Cr2O7, times molarity with the volume
(f) find the mole ratio between TeO2 and Cr2O7 (3:1) and the molar mass of TeO2 and find the
mass of tellurite
AP Chemistry 2011 free response question
(a) (i) M1V1 = M2V2; should only have 1 significant figure (6 M)
(ii) lab procedures: wear safety goggles and rubber gloves. Measure 19mL of 16 M HNO3 using
100 mL graduated cylinder. Measure 31 mL of H2O (50-19) using 100 mL graduated cylinder. Mix
those in 100 mL beaker
(iii) graduated cylinder already has precision in volume measurement (19 mL and 39 mL)
(iv) NaHCO3 will be used since the HCO3^- ion will react as a base to neutralize the HNO3
Annotated Answers:
(b) third weighing – filter crucible to find the mass of the AgCl precipitate, then use the molar mass
to find the number of moles of AgCl precipitate
(c) find the mass of Ag from the number of moles of AgCl precipitate. Silver mass / total mass x
100% to find the mass percent of Ag