Practice Multiple Choice
1.
BrO3- + 5 Br- + 6 H+  3 Br2 + 3 H2O
How many moles of Br2 can be produced when 25 mL of 0.20 M BrO3- is mixed with 30 mL of 0.45 M Br-?
(A) 0.0050 (B) 0.0081 (C) 0.014 (D) 0.015
.025 L x 0.20 mol BrO3-/L x 3 mol Br2/1 mol BrO3- = 0.015
B
.030 L x 0.45 mol Br-/L x 3 mol Br2/5 mol Br- = 0.0081
2.
3 Ag + 4 HNO3  3 AgNO3 + NO + 2 H2O
If 0.10 mole of silver is added to 10 mL of 6.0 M nitric acid, the number of moles of NO gas that can be formed is
(A) 0.015 (B) 0.020 (C) 0.033 (D) 0.045
0.010 L x 6 mol HNO3/L x 1 mol NO/4 mol HNO3 = 0.015
A
0.10 mol Ag x 1 mol/3 mol Ag = 0.033  0.015 mol NO
3.
2 KOH + SO2  K2SO3 + H2O
What mass of SO2 reacts with 1.0 L of 0.25-M KOH?
(A) 4.0 g
(B) 8.0 g
(C) 16 g
(D) 20. g
1.0 L x 0.25 mol KOH x 1 mol SO2 x 64 g SO2 = 8.0 g
B
1L
2 mol KOH 1 mol SO2
4. How many moles Ba(NO3)2 should be added to 300. mL of 0.20 M Fe(NO3)3 to increase the [NO3-] to 1.0 M?
(A) 0.060 (B) 0.12
(C) 0.24
(D) 0.30
0.3 L x 0.2 mol Fe(NO3)3/1L = 0.06 mol x 3 = 0.18 mol
A
0.3 L x 1 mol NO3-/1L = 0.3 mol needed  0.06 mol Ba...
5.
What is the concentration of HC2H3O2 if it takes 32 mL of 0.50 M NaOH solution to neutralize 20. mL of the acid?
(A) 1.6 M (B) 0.80 M (C) 0.64 M (D) 0.60 M
.32 L NaOH x .5 mol/L x 1 mol HC.../1 mol = .16 mol HC...
B
0.16 mol HC.../0.020 L = 0.80 M
6.
2 HCl + Ba(OH)2  BaCl2 + 2 H2O
What volume of 1.5 M HCI neutralizes 25 mL of 1.2-M Ba(OH)2?
(A) 20. mL (B) 30. mL (C) 40. mL (D) 60. mL
.025 L x 1.2 mol Ba... x 2 mol HCl x
1L _ = 0.040 L
C
1L
1 mol Ba... 1.5 mol HCl
7.
(A)
(B)
(C)
(D)
C
Which is the net ionic equation for the precipitation reaction between Lead(II) nitrate and sodium phosphate?
Na+ + NO3-  NaNO3(s)
3 Pb(NO3)2(aq) + 2 Na3PO4(aq)  6 NaNO3(s) + Pb3(PO4)2(aq)
3 Pb2+ + 2 PO43-  Pb3(PO4)2(s)
3 Pb(NO3)2(aq) + 2 Na3PO4(aq)  Pb3(PO4)2(s) + 6 NaNO3(aq)
The precipitate is Pb3(PO4)2(s). Only Pb2+ and PO43- are reactants  3 Pb2+ + 2 PO43-  Pb3(PO4)2(s)
8.
0.20-L of 0.20 M K2CO3 is added to 0.30-L of 0.40 M Ba(NO3)2. Barium carbonate precipitates. The concentration of barium
ion, Ba2+, remaining in solution is
(A) 0.15 M (B) 0.16 M (C) 0.20 M (D) 0.24 M
B
.30 L x .40 mol/L = .12 mol Ba2+ – (.20 L x .20 mol/L = .040 mol CO32-) = 0.080 mol Ba2+/0.50 L = 0.16 M
9.
Na3PO4(aq) + 3 AgNO3(aq)  Ag3PO4(s) + 3 NaNO3(aq)
100 mL of 1.0 M Na3PO4 is mixed with 100 mL of 1.0 M AgNO3. Which is the order of increasing concentration of the ions
remaining in solution?
(A) [PO43-] < [NO3-] < [Na+]
(B) [PO43-] < [Na+] < [NO3-]
(C) [NO3-] < [PO43-] < [Na+]
(D) [Na+] < [NO3-] < [PO43-]
.1 L x 1 mol/L = .1 mol PO43- + .3 mol Na+
A
.1 L x 1 mol/L = .1 mol Ag+ + .1 mol NO3- (Ag+ = 0)
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10.
(A)
(B)
(C)
(D)
B
An aqueous solution contains Pb2+, Fe2+, and Cu2+ ions. Which will separate Pb2+ from the other ions at 25oC?
Adding dilute Na2S solution
Adding dilute HCI solution
Adding dilute NaOH solution
Adding dilute NH3 solution
Pb2+ is insoluble in Cl-, All ions are insoluble in S2- and OH-, Cu2+ forms complex with NH3, others form OH- ppt.
11.
Which is LEAST soluble in water?
(A) (NH4)2SO4
(B) KMnO4
(C) BaCO3
(D) Zn(NO3)2
C
All NH4+, K+ and NO3- salts are soluble. Most carbonates are insoluble including BaCO3.
Questions 12-13 refer to the following solid compounds.
(A) PbCl2 (B) CuSO4 (C) K2CrO4 (D) KCl
12.
Is yellow in aqueous solution
C
K2CrO4 (Used in the Qual analysis lab)
13.
A
Is white and insoluble in water
PbCl2 and KCl are white, but KCl is soluble whereas PbCl2 is insoluble.
Cl-, OH- and SO42- are added to three samples of a colorless, 0.10 M solution, respectively. Which cation could be
present in the solution if no change is observed in any sample?
(A) Ni2+
(B) Ag+
(C) Ba2+
(D) Na+
Ni2+ + 2 OH-  Ni(OH)2(s), Ag+ + Cl-  AgCl(s),
D
Ba2+ + SO42-  BaSO4(s)  only Na+ is spectator.
14.
15.
(A)
(B)
(C)
(D)
C
Which occurs when concentrated NH3 is mixed with 0.1 M Cu(NO3)2?
A dark red precipitate forms and settles out.
Separate layers of immiscible liquids form.
The color turns from light blue to dark blue.
Bubbles of ammonia gas form.
Cu2+ + 4 NH3  Cu(NH3)42+
light blue
dark blue
16.
A solution containing cations is treated with 0.1 M HCl, forming a white precipitate, which is centrifuged while warm.
0.1 M K2CrO4 is added to supernatant and a bright yellow precipitate forms. The white precipitate dissolves in
ammonia solution. What ions are present?
(A) Ag+ and Hg22+
(B) Ag+ and Pb2+
2+
2+
(C) Hg2 and Pb
(D) Pb2+, Ag+ and Hg22+
B
Pb2+ is soluble in hot and forms a precipitate with CrO42- (PbCrO4)  Pb2+. Ag+ is soluble in NH3
17.
(A) H2S
C
What is the correct formula for sulfurous acid?
(B) HSO3 (C) H2SO3 (D) H2SO4
Without hydro  oxyacid, "ous" ending "ite" anion = H2SO3
18.
In which species does sulfur have the same oxidation number as it does in H 2SO4?
(A) H2SO3 (B) S2O32- (C) S2(D) SO2Cl2
D
H2SO4: 2+S–8=0 (S=6), H2SO3: 2+S–
2O3
2-:
2S–6=-2 (S=4), S2-: S=-2, SO2Cl2: S–4 –2=0 (S=6)
19.
2 HClO + 3 O2  2 HClO4
The oxidation number of chlorine changes from
(A) -1 to +3 (B) -1 to +5 (C) +1 to +7 (D) +3 to +7
ox.# 1 1 -2
0
1 7 -2
C
2 HClO + 3 O2  2 HClO4
2
20.
Which will generate H2(g) when added to 1 M HCl?
(A) CuS
(B) Zn
(C) CaCO3 (D) Mg(OH)
Zn + 2 H+  Zn2+ H2(g)
B
(CuS  H2S, CaCO3  CO2, Mg(OH)2  H2O)
21.
3 Cu + 8 H+ + 2 NO3-  3 Cu2+ + 2 NO + 4 H2O
Which statements about the reaction are true?
I. Cu acts as an oxidizing agent.
II. Nitrogen's oxidation state changes from +5 to +2.
III. Hydrogen ions are oxidized to form H2O.
(A) I only
(B) II only (C) III only (D) I and II
ox # 0
1
5 -2
2
2 -2
1 -2
B
3 Cu + 8 H+ + 2 NO3-  3 Cu2+ + 2 NO + 4 H2O
22.
10 HI + 2 KMnO4 + 3 H2SO4  5 I2 + 2 MnSO4 + K2SO4 + 8 H2O
How many moles of HI are used to produce 2.5 mol of I2?
(A) 5.0
(B) 8.0
(C) 10.
(D) 12
2.5 mol I2 x 10 mol HI = 5 mol HI
A
5 mol I2
23.
2 H2O + 4 MnO4- + 3 CIO2-  4 MnO2 + 3 CIO4- + 4 OHHow many moles of ClO2- react with 0.20 L of 0.20 M MnO4-?
(A) 0.030 (B) 0.053 (C) 0.075 (D) 0.13
0.20 L x 0.20 mol MnO4- x 3 mol ClO2- = 0.030 mol
A
1L
4 mol MnO424.
5 Fe2+ + MnO4- + 8 H+  5 Fe3+ + Mn2+ + 4 H2O
25.0 mL of an acidified Fe2+ solution requires 14.0 mL of 0.10-M MnO4- solution to reach the equivalence point. The concentration
of Fe2+ in the original solution is
(A) 0.10 M (B) 0.56 M (C) 0.28 M (D) 0.14 M
0.014 L x 0.10 mol MnO4- x 5 mol Fe3+ = 0.0070 mol Fe3+
C
1L
1 mol MnO40.025 L
Questions 25-27 refer to the following reactions.
(A) 2 Mg(s) + O2(g)  2 MgO(s)
(B) Pb2+(aq) + CrO42-(aq)  PbCrO4(s)
(C) 2 H2O(g)  2 H2(g) + O2(g)
(D) Ag+(aq) + 2 NH3(aq)  [Ag(NH3)2]+(aq)
25.
B
26.
D
27.
A
A precipitation reaction
The solid product is made from ions.
A reaction that produces a coordination complex
The product contains anions attached to a metal cation, but the product has an overall charge.
A synthesis reaction that is also oxidation-reduction.
The compound MgO is made from its elements when Mg is oxidized and O is reduced.
Questions 28-31 refer to the reactions represented below.
(A) S8(s) + 8 O2(g)  8 SO2(g)
(B) 3 Br2(aq) + 6 OH-  5 Br- + BrO3- + 3 H2O(l)
(C) Ca2+ + SO42-  CaSO4(s)
(D) PtCI4(s) + 2 CI-  [PtCI6]228.
C
29.
D
A precipitation reaction
The solid product is made from ions.
A reaction that produces a coordination complex
The product contains anions attached to a metal cation, but the product has an overall charge.
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30.
B
A reaction in which the same reactant undergoes both oxidation and reduction
The bromine (ox # = 0) becomes Br- (ox # = -1) and BrO3- (ox # = +5).
31.
A
A combustion reaction
Combustion reactions are also redox reactions, where the oxidizing agent is oxygen gas.
Questions 32-35 refer to the chemical reactions represented below.
(A) 2 HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + H2O
(B) Ba2+ + SO42-  BaSO4(s)
(C) Zn(OH)2(s) + 2 OH-  [Zn(OH)4]2(D) 2 K(s) + Br2(l)  2 KBr(s)
(E) N2O4(g)  2 NO2(g)
32.
D
An oxidation-reduction reaction
K is oxidized and Br is reduced.
33.
B
A precipitation reaction
The solid product is made from ions.
34.
C
A reaction in which a coordination complex is formed
The product contains anions attached to a metal cation, but the product has an overall charge.
35.
A
An acid-base reaction
H+
and OH- combine to form H2O (counterions form the aqueous salt).
Practice Free Response
1.
For each of the following four reactions, write a balanced equation in part (1) and answer the question in part (2).
a. Solid lithium oxide is added to distilled water.
(1) Balanced equation:
Li2O(s) + H2O(l)  2 Li2+(aq) + 2 OH-(aq)
(2) If a few drops of phenolphthalein are added to the resulting solution, what would be observed? Explain.
The phenolphthalein would turn pink because of the presence of OH- in solution making it basic.
b.
A barium nitrate solution and a potassium fluoride solution are combined and a precipitate forms.
(1) Balanced equation:
Ba2+(aq) + 2 F-(aq)  BaF2(s)
(2) If equimolar amounts of barium nitrate and potassium fluoride are combined, which reactant, if any, is the limiting
reactant? Explain.
KF is the limiting reactant because it takes two moles of KF to react with 1 mol of Ba(NO3)2  half of the Ba(NO3)2 will
remain after all of the KF has reacted.
c.
Solid strontium hydroxide is added to nitric acid solution.
(1) Balanced equation:
Sr(OH)2(s) + 2 H+(aq)  2 H2O(l) + Sr2+(aq)
(2) How many moles of strontium hydroxide would react completely with 500. mL of 0.40 M nitric acid?
0.500 L H+ x 0.40 mol H+/1 L x 1 mol OH-/1 mol H+ x 1 mol Sr(OH)2/1 mol OH- = 0.100 mol Sr(OH)2
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d.
A solution containing silver(I) ion (an oxidizing agent) is mixed with a solution containing iron(II) ion (a reducing agent).
(1) Balanced equation:
Ag+(aq) + Fe2+(aq)  Ag(s) + Fe3+(aq)
(2) If the contents of the reaction mixture described above are filtered, what substance(s), if any, would remain on the filter
paper?
The precipitated solid silver remains on the filter paper.
2.
1.625 g of pure acetylsalicylic acid reacts with NaOH. The reaction requires 88.43 mL of 0.102 M NaOH.
a. Calculate the molar mass of the acid. (it takes one mole of acid to react with each mole of NaOH)
0.08843 L OH- x 0.102 mol OH- x 1 mol HA = 0.00902 mol
1L
1 mol OH1.625 g/0.00902 mol = 180. g/mol
b.
The NaOH buret was rinsed with distilled water resulting in the first few drops of NaOH to be more dilute. Would the calculated
molar mass of acid be larger, smaller or unaffected by this variations in procedure. Briefly explain your reasoning.
It will take more drops of NaOH to reach equivalence  the calculated moles of acid would be too high and the molar
mass would be too low.
3.
5 Fe2+ + MnO4- + 8 H+  5 Fe3+ + Mn2+ + 4 H2O
0.500-g of iron(II) compound is dissolved in distilled water, acidified, and then titrated with 13.5 mL of 0.0200 M KMnO4 to reach the
end point. Determine the following
a. The number of moles of MnO4- used.
0.0135 L x 0.0200 mol MnO4- = 2.70 x 10-4 mol MnO41 L MnO4b. The number of moles of iron in the sample.
2.70 x 10-4 mol MnO4- x 5 mol Fe2+ = 1.35 x 10-3 mol Fe2+
1 mol MnO4c. The mass of iron in the sample, in grams.
1.35 x 10-3 mol Fe2+ x 55.8 g Fe2+ = 0.0753 g Fe2+
1 mol Fe2+
d.
The mass percent of iron in the compound.
0.0753 g Fe2+/0.500 g x 100 = 15.1% Fe2+
4.
Answer the following questions about BeC2O4(s) and its hydrate.
a. Calculate the mass percent of carbon in the hydrated form of the solid that has the formula BeC 2O4•3 H2O
MM = 9.01 + 2(12.0) + 4(16.0) + 3(18.0) = 151 g/mol
% C = [2(12.0)/151] x 100 = 15.9 %
b. When heated to 220.oC, BeC2O4•3 H2O(s) dehydrates completely as represented below.
BeC2O4•3 H2O(s)  BeC2O4(s) + 3 H2O(g)
If 3.21 g of BeC2O4•3 H2O(s) is heated to 220.oC, calculate
(1) the mass of BeC2O4(s) formed
3.21 g BeC2H4•3 H2O x 97.0 g BeC2O4 = 2.06 g BeC2O4
151 g BeC2O4•3 H2O
(2) the volume of the H2O(g) released, measured at 220.oC and 735 mm Hg.
3.21 g BeC2H4•3 H2O x 1 mol x 3 mol H2O = 0.0638 mol
151 g 1 mol Be...
PV = nRT
V = nRT = (0.0638 mol)(0.0821 atm•L/mol•K)(493 K)
P
(735/760 atm)
V = 2.67 L
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c.
A 0.345 g sample of anhydrous BeC2O4, which contains an inert impurity, was dissolved in sufficient water to produce 100. mL
of solution. A 20.0 mL portion of the solution was titrated with KMnO4(aq). The balanced equation for the reaction that
occurred is as follows.
16 H+ + 2 MnO4- + 5 C2O42-  2 Mn2+ + 10 CO2(g) + 8 H2O
The volume of 0.0150 M KMnO4(aq) required to reach the equivalence point was 17.80 mL.
(1) Identify the reducing agent in the titration reaction.
C2O42- is the reducing agent (the oxidation number for C in C2O42- is +3 and in CO2 is +4, which is an increase  C2O42is acting as an reducing agent)
(2) For the titration at the equivalence point, calculate the number of moles of each of the following that reacted.

MnO40.01780 L MnO4- 0.0150 mol = 2.67 x 10-4 mol MnO41L

C2O422.67 x 10-4 mol MnO4- x 5 mol C2O42- = 6.68 x 10-4 mol C2O422 mol MnO4(3) Calculate the total number of moles of C2O42- that were present in the 100. mL of prepared solution.
100 mL x 6.68 x 10-4 mol C2O42- = 3.34 x 10-3 mol C2O4220 mL
(4) Calculate the mass percent of BeC2O4(s) in the impure 0.345 g sample.
.00334 mol C2O42- x 1 mol BeC2O4 x 97.0 g =.324 g BeC2O4
1 mol C2O42- 1 mol Be...
0.324 g/0.345 g x 100 = 93.9 %
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