Chemistry for today general organic and biochemistry 8th edition seager solutions manual 1 by james.simmons193

Chapter 5: Chemical Reactions

Solution Manual for Chemistry for Today General Organic and Biochemistry Hybrid 8th Edition by Seager Slabaugh

ISBN 1285185978 9781285185972

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CHAPTER OUTLINE

5 1 Chemical Equations

5 2 Types of Reactions

5.3 Redox Reactions

5 4 Decomposition Reactions

5 5 Combination Reactions

5 6 Replacement Reactions

5.7 Ionic Equations

5 8 Energy and Reactions

LEARNING OBJECTIVES/ASSESSMENT

5 9 The Mole and Chemical Equations

5.10 The Limiting Reactant

5 11 Reaction Yields

When you have completed your study of this chapter, you should be able to:

1. Identify the reactants and products in written reaction equations, and balance the equations by inspection (Section 5.1; Exercises 5.2 and 5 6)

2 Assign oxidation numbers to elements in chemical formulas, and identify the oxidizing and reducing agents in redox reactions. (Section 5 3; Exercises 5.10 and 5 15)

3 Classify reactions into the categories of redox or nonredox, then into the categories of decomposition, combination, single replacement, or double replacement. (Sections 5.4, 5.5, and 5.6; Exercise 5.20)

4 Write molecular equations in total ionic and net ionic forms. (Section 5 7; Exercise 5 30 a, b, & c)

5 Classify reactions as exothermic or endothermic. (Section 5 8; Exercise 5 34)

6. Use the mole concept to do calculations based on chemical reaction equations. (Section 5.9; Exercise 5.42)

7. Use the mole concept to do calculations based on the limiting-reactant principle. (Section 5.10; Exercise 5.52)

8. Use the mole concept to do percentage-yield calculations. (Section 5.11; Exercise 5.56)

LECTURE HINTS AND SUGGESTIONS

1. Do not get bogged down explaining and defining oxidation numbers. Oxidation numbers can easily be overemphasized. Their main roles are in recognizing redox reactions and in balancing redox equations. Focus on those roles as the purpose for learning the rules of oxidation numbers. Make certain that students understand that oxidation numbers are not charges.

2. Use some simple redox reactions and relate the oxidation number change with the loss and gain of electrons. For example, in the reaction, 2Na + Cl2 → 2NaCl, each sodium atom loses one electron and its oxidation number goes from 0 to + 1; and each chlorine atom gains an electron and its oxidation number goes from 0 to -1.

3. Sometimes a concrete example of a limiting reactant helps the student to visualize that concept. One can obtain individual flavors of Jelly Belly ® jelly beans from most candy stores. The company also publishes various combinations that result in new flavors. One such flavor is ‘Orange Dream Bars’ produced by eating 2 orange sherbet and one cream soda at the same time. Purchase a pound of each flavor and have your students produce ‘Orange Dream Bars’ – by eating them. At some point, they will run out of orange sherbet – showing it to be the limiting reactant. Other combinations are listed at www.jellybelly.com.

4. Many double replacement reactions are easy to demonstrate in class. Simply mix two solutions, which will provide two ions that form a water insoluble product; for example, sodium carbonate and calcium chloride or barium chloride and sodium sulfate or silver nitrate and potassium chromate. Explain that when two ions can combine to form a water insoluble product, then a precipitate will always form upon mixing the ions. Try to pick examples that have some relevance such as in these cases where the formation of calcium carbonate would illustrate what is meant by hard water and where the insolubility of barium sulfate is related to its usefulness as a radio-opaque diagnostic material. The silver chromate is a chemical precursor to modern photography that has been important in neuroscience, as it is used in the “Golgi method” of staining neurons for microscopy. The reaction that produces silver chromate is also highly visible in a large lecture theater because the reaction involves a color change in addition to the formation of a precipitate.

SOLUTIONS FOR THE END OF CHAPTER EXERCISES CHEMICAL EQUATIONS (SECTION

copper, water 

Reactants Products

d. ethane + oxygen  carbon dioxide + water ethane, oxygen carbon dioxide, water

5.3 a. 4 Al (s) + 3 O2 (g)  2 Al2O3 (s) is consistent with the law of conservation of matter.

b. P4 (s) + O2 (g) P4O10 (s) is not consistent with the law of conservation of matter because the number of oxygen atoms is not the same on both sides of the reaction.

c. 3.20 g oxygen + 3.21 g sulfur  6.41 g sulfur dioxide is consistent with the law of conservation of matter.

Notice, the number of moles of oxygen and sulfur are equal on both sides of the equation. 2

    

32.0gO

22 2

92 Chapter 5

5.1) 5.1 Reactants

a. BaO2 (s) + H2SO4 (l)  BaSO4 (s) + H2O2 (l) BaO2, H2SO4 BaSO4, H2O2 b. 2 H2O2 (aq)  2 H2O (l) + O2 (g) H2O2 H2O, O2

Products



c. methane + water

carbon monoxide + hydrogen methane, water carbon monoxide, hydrogen

d. copper(II) oxide + hydrogen  copper + water copper(II) oxide, hydrogen 5.2

H2, Cl2 HCl

a. H2 (g) + Cl

(g)  2 HCl (g)

b. 2 KClO3 (s)  2 KCl (s) + 3 O2 (g) KClO3 KCl, O2

c. magnesium oxide + carbon  magnesium + carbon monoxide magnesium oxide, carbon magnesium, carbon monoxide

1moleO

Reactants:3.20gO0.100molesO;

ZnS (s) + O2 (g)  ZnO (s) + SO2 (g) is not consistent with the law of conservation of matter because the reactant (left) side of the equation has two moles of oxygen atoms, while the product (right) side of the equation has three moles of oxygen atoms

b. Cl2 (aq) + 2 I- (aq) I2 (aq) + 2Cl- (aq) is consistent with the law of conservation of matter.

c. 1.50 g oxygen + 1.50 g carbon  2.80 g carbon monoxide is not consistent with the law of conservation of matter because the mass of the reactants is 3.00 g, while the mass of the products is only 2.80 g.

Notice, the number of moles of oxygen and carbon are not equal on both sides of the equation either.

This equation is balanced.

This equation is not balanced because the numbers of moles of oxygen atoms are not balanced.

This equation is balanced.

Chemical Reactions 93 2 2 22 2 2 22 1moleS 3.21gS0.100molesS 32.1gS 1moleSO2molesO Products:6.41gSO0.200molesO; 64.1gSO1moleSO 1moleSO1moleS 6.41gSO0.100molesS 64.1gSO1moleSO               

CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)

with

5.4 a.

is consistent

the law of conservation of matter.

2 22 2 1moleO Reactants:1.50gO0.0469molesO; 32.0gO      1moleC 1.50gC0.125molesC 12.0gC 1moleCO1moleO Products:2.80gCO0.100moleO; 28.0gCO1moleCO 1moleCO1moleC 2.80gCO0.100moleC 28.0gCO1moleCO                d. 2 C2H6 (g) + 7 O2 (g)  4 CO2 (g) + 6 H2O (g) is consistent with the law of conservation of matter. 5.5 a. H2S (aq) + I2 (aq) → 2 HI (aq) + S (s) Elements H S I

Reactant 2 1 2 Product 2 1 2 b. KClO3 (s) → KCl (s) + O2 (g) Elements K Cl O

Reactant 1 1 3 Product 1 1 2 c. SO2 (g) + H2O (l) → H2SO3 (aq) Elements S O H

Reactant 1 3 2 Product 1 3 2

REDOX REACTIONS (SECTION 5.3)

This

94 Chapter

Ba(ClO3)2 (aq) + H2SO4 (aq) → 2HClO3 (aq) + BaSO4 (s) Elements Ba Cl O H S This equation is balanced Reactant 1 2 10 2 1 Product 1 2 10 2 1 5.6 a. Ag (s) + Cu(NO3)2 (aq) → Cu (s) + AgNO3 (aq) Elements Ag Cu N O

equation is not balanced because the number of moles of nitrogen and oxygen are not balanced. Reactant 1 1 2 6 Product 1 1 1 3 b. 2 N2O (g) + 3 O2 (g) → 4 NO2 (g) Elements N O

5 d.

This

equation is balanced. Reactant 4 8 Product 4 8

Mg (s) + O2 (g) → 2 MgO (s) Elements Mg O

equation

moles of magnesium

Reactant 1 2 Product 2 2 d. H2SO4 (aq) + Ca(OH)2 (aq) → CaSO4 (s) + 2 H2O (l) Elements H S O Ca This equation is balanced. Reactant 4 1 6 1 Product 4 1 6 1 5.7 a. Cl2 (aq) + NaBr (aq)→ NaCl (aq) + Br2 (aq) Cl2 (aq) + 2 NaBr (aq)→ 2 NaCl (aq) + Br2 (aq) b. CaF2 (s) + H2SO4 (aq) → CaSO4 (s) + HF (g) CaF2 (s) + H2SO4 (aq) → CaSO4 (s) + 2 HF (g) c. Cl2 (g) + NaOH (aq) → NaOCl (aq) + NaCl (aq) + H2O (l) Cl2 (g) + 2 NaOH (aq) → NaOCl (aq) + NaCl (aq) + H2O (l) d. KClO3 (s) → KClO4 (s) + KCl (s) 4 KClO3 (s) → 3 KClO4 (s) + KCl (s) e. dinitrogen monoxide → nitrogen + oxygen 2 N2O → 2 N2 + O2 f. dinitrogen pentoxide → nitrogen dioxide + oxygen 2 N2O5 → 4 NO2 + O2 g. P4O10 (s) + H2O (l) → H4P2O7 (aq) P4O10 (s) + 4 H2O (l) → 2 H4P2O7 (aq) h. CaCO3 (s) + HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g) CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g) 5.8 a. KClO3 (s) → KCl (s) + O2 (g) 2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) b. C2H6 (g) + O2 (g) → CO2(g) + H2O (l) 2 C2H6 (g) + 7 O2 (g) → 4 CO2(g) + 6 H2O (l) c. nitrogen + oxygen → dinitrogen pentoxide 2 N2 (g) + 5 O2 (g) → 2 N2O5 (g) d. MgCl2 (s) + H2O (g) →MgO (s) + HCl (g) MgCl2 (s) + H2O (g) → MgO (s) + 2 HCl (g) e. CaH2 (s) + H2O (l) → Ca(OH)2 (s) + H2 (g) CaH2 (s) + 2 H2O (l) → Ca(OH)2 (s) + 2 H2 (g) f. Al (s) + Fe2O3 (s) → Al2O3 (s) + Fe (s) 2 Al (s) + Fe2O3 (s) → Al2O3 (s) + 2 Fe (s) g. aluminum + bromine → aluminum bromide 2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s) h. Hg2(NO3)2 (aq) + NaCl (aq) → Hg2Cl2 (s) + NaNO3 (aq) Hg2(NO3)2 (aq) + 2 NaCl (aq) → Hg2Cl2 (s) + 2 NaNO3 (aq)

is not balanced because the number of

is not balanced

Chemical Reactions 95 5.9 Calculations O.N. a. Cl2O5     25 Cl?O2ClO0 2?5(2)0?5 +5 b. KClO4   4 K1Cl?O2KClO0 1?4(2)0?7 +7 c. Ba2+ +2 d. F2 0 e. H4P2O7         427 H1P?O2HPO0 412?720?5 +5 f. H2S     2 H1S?HS0 21?0?2 -2 5.10 Calculations O.N. a. HClO   H1Cl?O2HClO0 1?(2)0?1   +1 b. NaNO2       2 Na1N?O2NaNO0 1?220?3 +3 c. N2 0 d. Ca(ClO)2         2 Ca2Cl?O2NO0 22?20?1 +1 e. Al3+ +3 f. N2O5     2 N?O2NO0 2?5(2)0?5 +5 5.11 Oxidation Numbers Highest O.N. a. N2O5 N = +5, O = -2 nitrogen b. KHCO3 K = +1, H = +1, C = +4, O = -2 carbon c. NaOCl Na = +1, O = -2, Cl = +1 sodium and chlorine d. NaNO3 Na = +1, N = +5, O = -2 nitrogen e. HClO4 H = +1, Cl = +7, O = -2 chlorine f. Ca(NO3)2 Ca = +2, N = +5, O = -2 nitrogen 5.12 Oxidation Numbers Highest O.N. a. Na2Cr2O7 Na = +1, Cr = +6, O = -2 chromium b. K2S2O3 K = +1, S = +2, O = -2 sulfur c. HNO3 H = +1, N = +5, O = -2 nitrogen d. P2O5 P = +5, O = -2 phosphorus e. Mg(ClO4)2 Mg = +2, Cl = +7, O = -2 chlorine f. HClO2 H = +1, Cl = +3, O = -2 chlorine 5.13 O.N. Change Classification a. 2 Mg (s) + O2 (g) → 2 MgO (s) 0→+2 lost e- oxidized b. CuO (s) + H2 (g) → Cu (s) + H2O (g) +2→0 gained e- reduced c. Ag+ (aq) + Cl- (aq) → AgCl (s) +1→+1 no change neither

Elements H O C

Reactant +1 -2 -4

Product 0 -2 +2

c. CuO (s) + H2 (g)  Cu (s) + H2O (g)

Elements Cu O H

Reactant +2 -2 0

Product 0 -2 +1

d. B2O3 (s) + 3 Mg (s) → 2 B (s) + 3 MgO (s)

Elements B O Mg

Reactant +3 -2 0

Product 0 -2 +2

e. Fe2O3 (s) + CO (g)  2 FeO (s) + CO2 (g)

Elements Fe O C

Reactant +3 -2 +2

Product +2 -2 +4

The oxidizing agent is H2O. The reducing agent is CH4

The oxidizing agent is CuO The reducing agent is H2

The oxidizing agent is B2O3 The reducing agent is Mg.

The oxidizing agent is Fe2O3 The reducing agent is CO

f. Cr2O7 2- (aq) + 2 H+ (aq) + 3 Mn2+ (aq) → 2 Cr3+ (aq) + 3 MnO2 (s) + H2O (l)

Elements Cr O H Mn

Reactant +6 -2 +1 +2

Product +3 -2 +1 +4

5.16 a. 2 Cu (s) + O2 (g) → 2 CuO (s)

Elements Cu O

Reactant 0 0

Product +2 -2

The oxidizing agent is Cr2O7 2The reducing agent is Mn2+

The oxidizing agent is O2 The reducing agent is Cu

b. Cl2 (aq) + 2 KI (aq)  2 KCl (aq) + I2 (aq)

96 Chapter 5 d. BaCl2(aq) + H2SO4 (aq) → BaSO4 (s) + 2 HCl (aq) +2→+2 no change neither e. Zn (s) + 2 H+ (aq) → Zn2+ (aq) + H2 (g) 0→+2 lost e- oxidized 5.14 O.N. Change Classification a. 4 Al (s) + 3 O2 (aq) → 2 Al2O3 (s) 0 → +3 lost e- oxidized b. SO2 (g) + H2O (l)  H2SO3 (aq) +1 → +1 no change neither c. 2 KClO3 (s)  2 KCl (s) + 3 O2 (g) +5 → -1 gained e- reduced d. 2 CO (g) + O2 (g)  2 CO2 (g) +2 → +4 lost e- oxidized e. 2 Na (s) + 2 H2O (l)  2 NaOH (aq) + H2 (g) 0 → +1 lost e- oxidized 5.15 a. H2 (g) + Cl2 (g) → 2 HCl (g) Elements H Cl The oxidizing agent is Cl2. The reducing agent is H2 Reactant 0 0 Product +1 -1 b. H2O (g) + CH

4 (g)  CO (g) + 3 H2 (g)

The

→ 6 Ag (s) + Al2S3 (s)

The oxidizing agent is Ag2S because the oxidation number for the silver changes from +1 to 0. The reducing agent is Al because the oxidation number for the aluminum changes from 0 to +3.

5.18 6 NaOH (aq) + 2 Al (s) → 3 H2 (g) + 2 Na3AlO3 (aq) + heat

The oxidizing agent is NaOH because the oxidation number for the hydrogen changes from +1 to 0. The reducing agent is Al because the oxidation number for the aluminum changes from 0 to +3.

5.19 N2 (g) + 3 H2 (g) → 2 NH3 (g)

The oxidizing agent is N2 because the oxidation number for the nitrogen changes from 0 to -3. The reducing agent is H2 because the oxidation number for the hydrogen changes from 0 to +1.

DECOMPOSITION, COMBINATION, AND REPACEMENT REACTIONS (SECTION 5.4-5.6) 

b. Ca (s) + 2

+ CO2 (g)

O (l) → Ca(OH)2 (s) + H2 (g)

decomposition

+1 0 redox: single-replacement

(aq) → BaSO4 (s) + 2 HCl (aq) nonredox: double-replacement d.

c. BaCl

combination

Chemical Reactions 97 Elements Cl K I

oxidizing

2 The reducing

KI. Reactant 0 +1 -1 Product -1 +1 0

3 MnO2 (s) + 4 Al (s)  2 Al2O3 (s) + 3 Mn (s) Elements

O Al The oxidizing

2. The reducing agent is Al. Reactant +4 -2 0 Product 0 -2 +3 d. 2 H+ (aq) + 3 SO3 2- (aq) + 2 NO3 - (aq) → 2 NO (g) + H2O (l)+ 3 SO4 2- (aq) Elements H S O N

oxidizing

3. The reducing agent is SO3 2Reactant +1 +4 -2 +5 Product +1 +6 -2 +2 e. Mg (s) + 2 HCl (aq)  MgCl2 (aq) + H2 (g) Elements

H Cl

agent is Cl

agent is

agent is MnO

The

agent is NO

oxidizing

The reducing

Reactant 0 +1 -1 Product +2 0 -1 f. 4 NO2 (g) + O2 (g)  2 N2O5 (g) Elements N O O

oxidizing

The reducing agent is NO2 Reactant +4 -2 0 Product +5 -2 -2 5.17 3 Ag2S (s) + 2 Al (s)

agent is HCl.

agent is Mg.

The

agent is O2

5.20 a. K2CO3 (s) → K2O (s)

nonredox:

0 +1 -2

-2

nonredox:

(aq) +

(g) +

O (l) →

SO3 (aq)

98 Chapter 5

2 NO (g) + O2 (g) → 2 NO2 (g) +2 -2 0 +4 -2 redox: combination

2 Zn (s) + O2 (g) → 2 ZnO (s) 0 0 +2 -2 redox: combination

a. N2O5 (g) + H2O (l) → 2 HNO3 (aq) nonredox: combination

Cr2O3 (s) + 2 Al (s) → 2 Cr(s) + Al2O3 (s) +3 -2 0 0 +3 -2 redox: single-replacement

CaO (s) + SiO2 (s) → CaSiO3 (s) nonredox: combination

5.21

H2CO3 (aq) → CO2 (g) + H2O (l) nonredox: decomposition

PbCO3 (s) → PbO (s) + CO2 (g) nonredox: decomposition

Zn (s) + Cl2 (g) → ZnCl2 (s) 0 0 +2 -1 redox: combination

22 Heat 32322 2NaHCO(s)NaCO(s)HO(g)CO(g)  +1 +1 +4 -2 +1 +4 -2 +1 -2 +4 -2 This reaction is a nonredox decomposition reaction. 5.23 NaHCO3 (aq) + H+ (aq) → Na+ (aq) + H2O (l) + CO2 (g); This reaction is a nonredox reaction. +1 +1 +4 -2 +1 +1 +1 -2 +4 -2 5 24 CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g); This reaction is a redox reaction. -4 +1 0 +4 -2 +1 -2 5.25 2 H2O2 (aq) → 2 H2O (g) + O2 (g); This reaction is a redox reaction. +1 -1 +1 -2 0 5.26 Cl2 (aq) + H2O (l) → HOCl (aq) + HCl (aq); This is a redox reaction. 0 +1 -2 +1 -2 +1 +1 -1 5.27 Ca3(PO4)2 (s) + 4 H3PO4 (aq) → 3 Ca(H2PO4)2 (s) +2 +5 -2 +1 +5 -2 +2 +1 +5 -2 This reaction is a nonredox combination reaction. IONIC EQUATIONS (SECTION 5.7) 5.28 a. LiNO3 Li+, NO3 - d. KOH K+, OHb. Na2HPO4 2 Na+ , HPO4 2e. MgBr2 Mg2+ , 2 Brc. Ca(ClO3)2 Ca2+, 2 ClO3 - f. (NH4)2SO4 2 NH4 +, SO4 25.29 a. K2Cr2O7 2 K+, Cr2O7 2- d. Na3PO4 3 Na+, PO4 3b. H2SO4 2 H+, SO4 2e. NH4Cl NH4 +, Clc. NaH2PO4 Na+, H2PO4 - f. KMnO4 K+, MnO45.30  a. SO2 (aq) + H2O (l) → H2SO3 (aq) Total ionic equation: SO2 (aq) + H2O (l) → 2 H+ (aq) + SO3 2- (aq) Spectator ions: none Net ionic equation: SO2 (aq) + H2O (l) → 2 H+ (aq) + SO3 2- (aq)

Chemical Reactions 99  b. CuSO4 (aq) + Zn (s) → Cu (s) + ZnSO4 (aq) Total ionic equation: Cu2+(aq) + SO4 2-(aq) + Zn (s) → Cu (s) + Zn2+(aq) + SO4 2-(aq) Spectator ions: SO4 2-(aq) Net ionic equation: Cu2+(aq) + Zn (s) → Cu (s) + Zn2+(aq)  c. 2 KBr (aq) + 2 H2SO4 (aq) → Br2 (aq) + SO2 (aq) + K2SO4 (aq) + 2 H2O (l) Total ionic equation: 2 K+(aq) + 2 Br(aq) + 4 H+(aq) + 2 SO4 2-(aq) → Br2 (aq) + SO2 (aq) + 2 K+ (aq) + SO4 2- (aq) + 2 H2O (l) Spectator ions: 2 K+ (aq), SO4 2- (aq) Net ionic equation: 2 Br-(aq) + 4 H+(aq) + SO4 2-(aq) → Br2 (aq) + SO2 (aq) + 2 H2O (l) d. AgNO3 (aq) + NaOH (aq) → AgOH (s) + NaNO3 (aq) Total ionic equation: Ag+(aq) + NO3 -(aq) + Na+(aq) + OH-(aq) → AgOH (s) + Na+(aq) + NO3 -(aq) Spectator ions: Na+(aq), NO3 -(aq) Net ionic equation: Ag+(aq) + OH-(aq) → AgOH (s) e. BaCO3 (s) + 2 HNO3 (aq) → Ba(NO3)2 (aq) + CO2 (g) + H2O (l) Total ionic equation: BaCO3 (s) + 2 H+(aq) + 2 NO3 -(aq) → Ba2+(aq) + 2 NO3 -(aq) + CO2 (g) + H2O (l) Spectator ions: 2 NO3 -(aq) Net ionic equation: BaCO3 (s) + 2 H+(aq) → Ba2+(aq) + CO2 (g) + H2O (l) f. N2O5 (aq) + H2O (l) → 2 HNO3 (aq) Total ionic equation: N2O5 (aq) + H2O (l) → 2 H+(aq) + 2 NO3 -(aq) Spectator ions: none Net ionic equation: N2O5 (aq) + H2O (l) → 2 H+(aq) + 2 NO3 -(aq) 5.31 a. H2O (l) + Na2SO3 (aq) + SO2 (aq) → 2NaHSO3 (aq) Total ionic equation: H2O (l) + 2Na+(aq) + SO3 2-(aq) + SO2(aq) → 2Na+(aq) + 2 HSO3 - (aq) Spectator ions: 2Na+ (aq) Net ionic equation: H2O (l) + SO3 2- (aq) + SO2 (aq) → 2 HSO3 - (aq) b. 3 Cu (s) + 8 HNO3 (aq) → 3 Cu(NO3)2 (aq) + 2 NO (g) +4 H2O (l) Total ionic equation: 3 Cu (s) + 8H+ (aq) + 8NO3 - (aq) → 3 Cu2+ (aq) + 6 NO3 - (aq)+ 2 NO (g) 4 H2O (l) Spectator ions: 6 NO3 - (aq) Net ionic equation: 3 Cu (s) + 8H+ (aq) + 2NO3 - (aq) → 3 Cu2+ (aq) + 2 NO (g) 4 H2O (l) c. 2 HCl (aq) + CaO (s) → CaCl2 (aq) + H2O (l) Total ionic equation: 2 H+ (aq) + 2 Cl- (aq) + CaO (s) → Ca2+ (aq) + 2 Cl- (aq) + H2O (l)

5.33

The reactants and products are identical for all three net ionic equations.

Chapter

Spectator ions: 2 Cl- (aq) Net ionic equation: 2 H+ (aq) + CaO (s) → Ca2+ (aq) + H2O (l) d. CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + CO2 (aq) + H2O (l) Total ionic equation: CaCO3 (s) + 2H+(aq) + 2Cl-(aq) → Ca2+(aq) + 2Cl-(aq) + CO2 (aq) + H2O (l) Spectator ions: 2Cl-(aq) Net ionic equation: CaCO3 (s) + 2H+(aq) → Ca2+(aq) + CO2 (aq) + H2O (l) e. MnO2 (s) + 4 HCl (aq) → MnCl2 (aq) + Cl2 (aq) + 2 H2O (l) Total ionic equation: MnO2 (s) + 4H+(aq) + 4Cl-(aq) → Mn2+ (aq) + 2 Cl-(aq) + Cl2 (aq)+ 2 H2O (l) Spectator ions: 2 Cl-(aq) Net ionic equation: MnO2 (s) + 4H+(aq) + 2Cl-(aq) → Mn2+ (aq) + Cl2 (aq) + 2 H2O (l) f. 2 AgNO3 (aq) + Cu (s)→ Cu(NO3)2 (aq) + 2 Ag (s) Total ionic equation: 2 Ag+ (aq) + 2 NO3 - (aq) + Cu (s) → Cu2+ (aq) + 2 NO3 - (aq) + 2 Ag (s) Spectator ions: 2 NO3 - (aq) Net ionic equation: 2 Ag+ (aq) + Cu (s) → Cu2+ (aq) + 2 Ag (s) 5.32 a. HNO3 (aq) + KOH (aq) → KNO3 (aq) + H2O (l) Total ionic equation: H+(aq) + NO3 -(aq) + K+ (aq) + OH-(aq) → K+(aq) + NO3 -(aq) + H2O (l) Spectator ions: K+(aq), NO3 -(aq) Net ionic equation: H+ (aq) + OH-(aq) → H2O (l) b. H3PO4 (aq) + 3 NH4OH (aq) → (NH4)3PO4 (aq) + 3 H2O (l) Total ionic equation: 3 H+(aq) + PO4 3- (aq) + 3 NH4 + (aq) + 3 OH-(aq) → 3 NH4 + (aq) + PO4 3- (aq) + 3 H2O (l) Spectator ions: 3 NH4 + (aq), PO4 3- (aq) Net ionic equation: 3 H+ (aq) + 3 OH-(aq) → 3 H2O (l); H+ (aq) + OH-(aq) → H2O (l) c. HI (aq) + NaOH (aq) → NaI (aq) + H2O (l) Total ionic equation: H+(aq) + I-(aq) + Na+(aq) + OH-(aq) → Na+(aq) + I-(aq) + H2O (l) Spectator ions: Na+(aq), I-(aq) Net ionic equation: H+ (aq) + OH-(aq) → H2O (l)

100

a. HBr (aq) + RbOH (aq) → RbBr (aq) + H2O (l) Total ionic equation: H+(aq) + Br-(aq) + Rb+ (aq) + OH-(aq) → Rb+(aq) + Br-(aq) + H2O (l)

ionic

H+ (aq) + OH-(aq) → H2O (l)

Spectator ions: Br-(aq),

+ (aq) Net

equation:

The reactants and products are identical for all three net ionic equations.

ENERGY AND REACTIONS (SECTION 5.8)

5.34 The emergency hot pack becoming warm is an exothermic process because, as water is mixed with the solid, heat is released.

5.35 The evaporation process is endothermic. The liquid requires heat to be converted into a gas.

5.36 By insulating the ice, the heat in the food will not be able to travel into the ice as easily. It would be better if the ice were not wrapped in a thick insulating blanket. Wrapping the blanket around the ice and the food would be a better arrangement for keeping the food cold.

5.37 The evaporation process is endothermic. The liquid requires heat to be converted into a gas.

EQUATIONS (SECTION 5.9)

Statement1:1Sratom+2HOmolecules 1Sr(OH) formulaunit+1Hmolecule

Statement2:1moleSr+2molesHO 1moleSr(OH)1moleH

Statement3:6.0210Sratoms+12.010HOmolecules

Statement4:87.6gSr36.0gHO 121.6gSr(OH)2.0gH

Statement1:2HSmolecules+3O molecules 2HOmolecules+2SO molecules

Statement2:2molesHS+3molesO 2molesH O+2molesSO

Statement3:12.010HSmolecules+18.110Omolecules

Statement4:68.2gHS96.0gO 36.0gHO+128.2gSO

Chemical Reactions 101 b. H2SO4 (aq) + 2 LiOH (aq) → Li2SO4 (aq) + 2 H2O (l) Total ionic equation: 2 H+(aq) + SO4 2- (aq) + 2 Li+ (aq) + 2 OH-(aq) → 2 Li+ (aq) + SO4 2- (aq) + 2 H2O (l) Spectator ions: SO4 2- (aq), 2 Li+ (aq) Net ionic equation: 2 H+ (aq) + 2 OH-(aq) → 2 H2O (l); H+ (aq) + OH-(aq) → H2O (l) c. HCl (aq)

CsOH (aq) → CsCl (aq) + H2O (l) Total ionic equation: H+(aq) + Cl-(aq) + Cs+(aq) + OH-(aq) → Cs+(aq) + Cl-(aq) + H2O (l) Spectator ions: Cs+(aq), Cl-(aq) Net ionic equation: H+ (aq) + OH-(aq) → H2O (l)

MOLE AND

5.38 a. S (s) + O2 (g) → SO2 (g)     22 22 232323 22 22 Statement1:1Satom+1O molecule 1SO molecule Statement2:1moleS+1moleO 1moleSO Statement3:6.0210Satoms+6.0210Omolecules 6.0210SOmolecules Statement4:32.1gS32.0gO

Sr (s)

2 H2O (l) → Sr(OH)2 (s) + H2 (g)     222 222 2323 2 23 2

THE

CHEMICAL

64.1gSO b.

6.0210Sr(OH) fo   23 2 222 rmulaunits+6.0210Hmolecules

2 H2O (g) + 2 SO2

    2222 2222 2323 22 23 2

2 H2S (g) + 3 O2 (g) →

(g)

12.010HOm   23 2 2222 olecules+12.010SOmolecules

Statement1:4NH molecules+5O molecules 4NOmolecules+6HOmolecules

Statement2:4molesNH +5molesO 4molesNO+6molesHO

Statement3:24.110NH molecules+30.110Omolecules

Statement4:68.0gNH160.0gO 120.0gNO+108.0gHO

e. CaO (s) + 3 C (s) → CaC2 (s) + CO (g)

Statement1:1CaOformulaunit+3Catoms 1CaCformulaunit+1COmolecule

Statement2:1moleCaO+3molesC 1moleCaC+1moleCO

Statement3:6.0210CaOformulaunits+18.110 Catoms 6.0210CaCformulauni 

Statement4:56.1gCaO+36.0gC

23 2

ts6.0210COmolecules

102 Chapter 5

    322 322 2323 32 23

d. 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6

(g)

24.110NOmole   23

2 322 cules36.110HOmolecules

    2 2 2323 23 2



64.1gCaC+28.0gCO 5.39 a. N2 (g) + 3 H2 (g) → 2NH3 (g) 223 223 232424 223 2 Statement1:1N molecule+3H molecules 2NH molecules Statement2:1moleN +3molesH 2molesNH Statement3:6.0210N molecules+1.8110Hmolecules1.2010NHmolecules Statement4:28.0gN6     23.06gH 34.1gNH  b. 2 Na (s) + Cl2 (g) → 2 NaCl (s) 2 2 242324 2 Statement1:2Naatoms+1Cl molecule 2NaClformulaunits Statement2:2molesNa+1moleCl 2molesNaCl Statement3:1.2010Naatoms+6.0210Clmolecules 1.2010NaClformulaunits Statement4:46.0gNa71.     20gCl 117gNaCl  c. BaCl2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2 HCl (aq) 224 4 2244 2323 224 Statement1:1BaCl formulaunit+1HSO molecule 1BaSO formulaunit+2HClmolecules Statement2:1moleBaCl +1moleHSO 1moleBaSO+2molesHCl Statement3:6.0210BaCl formulaunits+6.02 10HSOmo    2324 4 2244 lecules 6.0210BaSO formulaunits+1.2010HClmolecules Statement4:208gBaCl98.1gHSO 233gBaSO+73.0gHCl    d. 2 H2O2 (aq) → 2 H2O (l) + O2 (g)      2222 2222 24 22 2423 22 22 Statement1:2HO molecules 2HOmolecules+1Omolecule Statement2:2molesHO 2molesHO+1moleO Statement3:1.2010HO molecules 1.2010HOmolecules6.0210Omolecules Statement4:68.0gHO 22 36.0gHO+32.0gO e. 2 C3H6 (g) + 9 O2 (g) → 6 CO2 (g) + 6 H2O (g)

Chemical Reactions 103     362 22 36222 2424 362 24 Statement1:2CH molecules+9O molecules 6CO molecules+6HOmolecules Statement2:2molesCH +9molesO 6molesCO+6molesHO Statement3:1.2010CH molecules+5.4210O molecules 3.6110C   24 22 36222 O molecules3.6110HOmolecules Statement4:84.1gCH +288gO 264gCO+108gH O 5.40 2 SO2 + O2 → 2 SO3 (g) 223 223 2323 22 23 3 2 Statement1:2SO molecules+1O molecule 2SO molecules Statement2:2molesSO +1moleO 2molesSO

12.010SOmolecules Statement4:128gSO      2332.0gO 160gSO  232323 222 232323 223 2323 32 23 2 Factors: 12.010SOmolecules6.0210Omolecules12.010SOmolecules 6.0210Omolecules12.010SOmolecules12.010S;;; Omolecules 12.010SOmolecules6.0210Omolecules ; 12.010SOmolecules12     23 3 2323 32 222323 223232 2 2 12.010SOmolecules ;; .010SOmolecules6.0210Omolecules 2molesSO1moleO2molesSO2molesSO1moleO2molesSO ;;;;;; 1moleO2molesSO2molesSO2molesSO2molesSO1moleO 128gSO32.0g ; 32.0gO   22323 23232 O128gSO160gSO32.0gO160gSO ;;;; 128gSO160gSO128gSO160gSO32.0gO

Statement3:12.010SOmolecules+6.0210Omolecules

5.41 2 SO2 + O2 → 2 SO3 (g) 223 223 2323 22 23 3 2 Statement1:2SO molecules+1O molecule 2SO molecules Statement2:2molesSO +1moleO 2molesSO Statement3:12.010SOmolecules+6.0210Omolecules 12.010SOmolecules Statement4:128gSO      2332.0gO 160gSO  2 32 3 128gSO 350gSO280gSO 160gSO      280 g SO2 must react to produce 350 g SO3. 5.42 CaCO3 (s) → CaO (s) + CO2 (g) 3 3 3 100gCaCO 500gCaO891.265597148gCaCO 56.1gCaO 891gCaCO       5.43 CaCO3 (s) → CaO (s) + CO2 (g)

This list does not include all possible factors.

104 Chapter 5 2 2 2 1moleCO 500gCaO8.91265597148molesCO 56.1gCaO 8.91molesCO       5.44 2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s) 2 2 2 479gBr 50.1gAl444.4055556gBr 54.0gAl 444gBrwithSF       5.45 2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s) 3 3 3 2molesAlBr 50.1gAl1.85555556molesAlBr 54.0gAl 1.86molesAlBrwithSF       5.46 3 3 3 233 533gAlBr 50.1gAl494.50555556gAlBr 54.0gAl 495gAlBrwithSF or 50.1gAl445gBr495.1gAlBr495gAlBr(SF)        5.47 3 Ag2S (s) + 2 Al (s) → 6 Ag (s) + Al2S3 (s) a. 2 2 2 54.0gAl 0.250gAgS0.01814516129gAl 744gAgS 1.8110gAlwithSF       b. 4 23 223 2 4 23 1moleAlS 0.250gAgS3.3602150510molesAlS 744gAgS 3.3610molesAlSwithSF       5.48 TiCl4 (s) + 2 Mg (s) → Ti (s) + 2 MgCl2 (s) 3 1000gTi48.6gMg 1.00kgTi1014.61377871gMg 1kgTi47.9gTi 1.0110gMgwithSF       5.49 C6H12O6 (aq) + 6 O2 (aq) → 6 CO2 (aq) + 6 H2O (l) a. 2 2 2 2 108gHO 1.00moleglucose108gHO 1moleglucose 1.0810gHO       b. 2 2 2 2 192gO 1.00moleglucose192gO 1moleglucose 1.9210gO       5.50 C6H12O2 (aq) + 8 O2 (aq) → 6 CO2 (aq) + 6 H2O (l)

5.51

CH4 (g) + 4 Cl2 (g) → CCl4 (l) + 4 HCl (g) Cl2 is the limiting reactant because the amount of Cl2 available will produce less CCl4 product than the amount of CH4 could produce, as shown by the calculations below:

b. Only 8.13 g CCl4 can be produced because Cl2 is the limiting reactant. Enough CH4 is present to make 38.5 g of CCl4, but not enough chlorine is present to react with the excess amount of CH4 See the calculation in part a.

N2 (g) + 2 O2 (g) → 2 NO2 (g) O2 is the limiting reactant as shown by the calculations below:

b. Only 71.9 g NO2 can be produced because O2 is the limiting reactant. Enough N2 is present to make 115 g of NO2, but not enough oxygen is present to react with the excess amount

Chemical Reactions 105      2 2 256gO 1.00molcaproicacid256gO 1molcaproicacid

5.10)

THE LIMITING REACTANT (SECTION

4 44 4 4 24 2 154gCCl 4.00gCH38.5gCCl 16.0gCH 154gCCl 15.0gCl8.13gCCl 284gCl          

5.52 a.

          2 22 2 2 22 2 2molesNO 1.25molesN2.50molesNO 1moleN 2molesNO 50.0gO1.56molesNO 64.00gO



of N2.           2 22 2 2 22 2 92.0gNO 50.0gO71.9gNO 64.0gO 92.0gNO 1.25molesN115gNO 1moleN 5.53 2 C2H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O (g) 1.54x103 g O2 is required to burn 500 g C2H2. The cylinder that contains 2000 g of oxygen is more than enough to burn all the acetylene.      3 2 2222 22 160gO 500gCH1538gO1.5410gO 52.0gCH 5 54 NH3 (aq) + CO2 (aq) + H2O (l) → NH4HCO3 (aq) 144 g NH4HCO

NH3

H2O

3 is the maximum mass in grams that can be produced by these reactants. CO2 is the limiting reactant because the amount given can produce less product than the amounts of either the NH3 or H2O given. The

and

are both excess reactants.

5.55 Cr2O3 (s) + 2 Al (s) → 2 Cr (s) + Al2O3 (s)

53.3 g of Al are required to completely react with 150 g Cr2O3. If 150 g Cr2O3 are reacted with 150 g Al, then the final mixture will contain the products Cr and Al2O3 as well as some unreacted

REACTION YIELDS (SECTION 5.11) 5 56

11.74g 10072.96%yield

5.57  14.37g 10081.88%yield

5 58 The theoretical yield is the sum of the masses of reactant A with reactant B because the amounts given are said to react exactly, which means both reactants will be completely used to make the product.

5.59 2 Ca (s) + O2 (g) →

5.60 2 HgO (s) → 2 Hg (l) + O2 (g)

ADDITIONAL EXERCISES

5.61 I would not expect argon to be involved in a redox reaction because it is a noble gas. Noble gases do not tend to form ions, and in order to participate in a redox reaction, argon would have to form ions.

5.62 barium chloride (aq) + sodium sulfate (aq)  sodium chloride (aq) + barium sulfate (s) Molecular equation:

106 Chapter 5                43 343 3 43 243 2 43 243 2 79.1gNHHCO 50.0gNH233gNHHCO 17.0gNH 79.1gNHHCO 80.0gCO144gNHHCO 44.0gCO 79.1gNHHCO 2.00molesHO158gNHHCO 1molesHO

     23 23 54.0gAl 150gCrO53.289gAl53.3gAl 152gCrO

Al.



16.09g

17.55g

9.04g



10072.5%yield 7.59g4.88g 

      112gCaO 2.00gCa2.7930gCaO 80.2gCa 2.26g 10080.9%yield 2.7930gCaO or     2.26g

112gCaO 2.00gCa 80.2gCa

2 CaO (s)

10080.9%yield

      402gHg 7.22gHgO6.6876gHg 434gHgO 5.95g

6.6876gHg or    

402gHg 7.22gHgO 434gHgO

10089.0%yield

5.95g 10089.0%yield

5.63 Sodium is the element with an electron configuration of 1

is the element with 15 protons in its nucleus.

5.65 The formula of the sample that was decomposed was

5.66 (c) is the balanced equation.

5.67 (d) is the balanced equation.

5.68 (c) is the balanced equation.

5.69 The coefficient before O2 is (b) 2 in the balanced equation, as shown below:

5.70 The coefficient before ammonia, NH3, is (b) 2 in the balanced equation, as shown below:

5.71 The oxidation number of sodium in NaI and NaNO

is (a) +1.

5.72 The oxidation number for nitrogen in HNO3 is (b) +5.

Chemical Reactions 107 BaCl2 (aq) + Na2SO4 (aq)  2 NaCl (aq) + BaSO4 (s) Total Ionic equation:  22 44Ba(aq)2Cl(aq)2Na(aq)SO(aq)2Na(aq)2Cl(aq) BaSO(s)

Ions:

Ionic equation: 22 44 Ba(aq)SO(aq)BaSO(s)

Spectator

2 Na+ (aq) and 2 Cl- (aq) Net

 3 3Na(s)P(s)NaP(s) 5.64        60 60 naturallyoccuring 60gFe 6.983g7.5gFe naturallyoccuring elementaliron 55.9g elementaliron

2O5

           24 23 1moleO 80.0gO5.00molesO 16.0gO 1moleN 1.2010atomsN1.99molesN 6.0210atomsN ALLIED HEALTH EXAM CONNECTION

Phosphorus

, dinitrogen pentoxide.

Mg (s) + 2 H2

Mg(OH)2

+ H2 (g)

O (g) →

(s)

2HgO → 2 Hg + O2

C12H22O11 → 12 C + 11 H2O

CH4 + 2 O2 (g) → CO2 + 2 H2O

N2 (g) + 3 H2 (g)   2 NH3 (g)

5.73 The oxidation number for sulfur in SO4 2- is (c) +6.

5.74 The oxidation number for sulfur in Na2S2O3 is (c) +2.

5.75 (b) Bromine is oxidized (Br→ Br2; losing electrons) and Mn in reduced (MnO4→Mn+2; gaining electrons).

5.76 (a) C2H4 (g) is oxidized and 3O2 (g) is reduced, in the equation, C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g). The oxygen changes from an oxidation number of 0 to -2 by gaining electrons (reduction). The carbon changes from an oxidation number of -2 to +4 by losing electrons (oxidation).

5.77 (b) Cu2+ (aq) is being oxidized because it is gaining electrons.

5.78 The answer is (b) oxidizing agent: 4 NO3 - (aq), reducing agent: Sn(s).

8 H+ (aq) + 6 Cl- (aq) + Sn (s) + 4 NO3 - (aq)→ SnCl6 2- (aq) + 4 NO2 (g) + 4 H2O (l) The tin changes from an oxidation number of 0 to +4 by losing electrons (oxidation); therefore, Sn is the reducing agent. The nitrogen changes from an oxidation number of +5 to +4 by gaining electrons (reduction); therefore, NO3 - is the oxidizing agent.

5.79 The chemical reaction: 2 Zn + 2 HCl → 2 ZnCl + H2 is an example of (d) single displacement.

5.80 This equation, Cl2 + 2e- → 2Cl-, is (a) a reduction reaction because Cl2 is gaining electrons.

5.81 The chemical reaction: 2 NaI + Cl2 → 2 NaCl + I2 is an example of a (c) single replacement reaction.

5.82 This equation, Mg(OH)2 + 2 HCl → MgCl2 + 2 H2O, is (d) double replacement reaction because the reactants are trading partners.

5.83 Exergonic reactions are not (c) uphill reactions.

5.84 The net ionic equation is (b) Ca2+ (aq) + 2 Cl- (aq) → CaCl2 (s).

108 Chapter 5        3 H1N?O2HNO0 1?320 ?5

     2 4 S?O2SO2 ?422 ?6

            223 Na1S?O2NaSO0 212?320 2?4 ?2

molecular equation: Ca(NO3)2 (aq) + 2 KCl (aq) → CaCl2 (s) + 2 KNO3 (aq)

total ionic equation:

Ca2+ (aq) + 2 NO3 - (aq) + 2 K+ (aq) + 2 Cl- (aq) → CaCl2 (s) + 2 K+ (aq) +2 NO3 - (aq)

spectator ions: 2 NO3 - (aq), 2 K+ (aq)

net ionic equation: Ca2+ (aq) + 2 Cl- (aq) → CaCl2 (s)

5.85 The chemical reaction: N2 (g) + 3 H2 (g)

2 NH3 (g) shows that (b) 3 moles of hydrogen gas are needed to react with one mole of nitrogen.

5.86 (c) 53.5 g NH4Cl can be produced.

5.87 The number of grams of hydrogen formed by the action of 6 grams of magnesium (atomic weight = 24) on an appropriate quantity of acid is (a) 0.5 g.

5.88 (d) 0.25 mole of CaCl2 is needed to form 0.5 mole of NaCl.

5.89 In the reaction: 4Al + 3O2 → 2Al2O3, (b) 36 grams O2 are needed to completely react with 1.5 moles of Al.

CHEMISTRY FOR THOUGHT

5.90 When a yield of more than 100% occurs for a compound prepared by precipitation from water solutions, it is likely that the “dry” compound is still moist and contains extra water mass.

Chemical Reactions 109

 

       34 3 33 3 34 NHHClNHCl 1moleNH1moleHCl 17gNH1.0moleNH36.5gHCl1.0moleHCl 17.0gNH36.5gHCl 17gNH36.5gHCl53.5gNHCl

       2 2 2 2 Mg2HMgH 2gH 6gMg0.5gH 24gMg

                2233 2 2 23 23 3 3 CaClNaCOCaCO2NaCl 1moleCaCl 0.5moleNaCl0.25moleCaCl 2molesNaCl 1moleNaCO 0.5moleNaCl0.25moleNaCO 2molesNaCl 1moleCaCO 0.5moleNaCl0.25moleCaCO 2molesNaCl

     2 2 96gO 1.5molesAl36gO 4molesAl

5.91 In rapid combustion, the oxidizing agent is oxygen and the reducing agent is the material that is being burned, like wood. A fire can be extinguished by making either the oxidizing agent or the reducing agent the limiting reactant. By removing the reducing agent (fuel) from the fire, the fire can be extinguished. This technique is often used in wild fire suppression. By removing the oxidizing agent (oxygen) from the fire, the fire can be extinguished. The technique of smothering the fire can be done with a fire blanket, aqueous foam, or inert gas. By enclosing the fire to quickly use all the available oxygen, the fire can be extinguished.

5.92 The bubbles formed during the reaction of hydrogen peroxide indicate that at least one of the products is a gas. Another material that might provide the enzyme catalyst needed for hydrogen peroxide decomposition is blood.

5.93 NaCl and AgNO3 are both water soluble. AgCl is not water soluble. If a few grams of solid AgCl were added to a test tube containing 3 mL of water and the mixture were shaken, the test tube would look cloudy white until the mixture had a chance to settle because AgCl is not water soluble. If a few grams of solid NaNO3 were added to a test tube containing 3 mL of water and the mixture were shaken, the solid should dissolve leaving a clear solution because NaNO3 is water soluble

+ 16 H+ (aq) → 4Cr3+ (aq) + 3 CH3COOH (aq) + 11 H2O (l) The initial color of the solution is orange-colored, but as it reacts with alcohol, the solution becomes pale violet. The darker violet the solution becomes, the more alcohol that was present in the breath.

5.95 The process of vegetables and fruits darkening when sliced is the result of a redox reaction. Placing these slices in water slows the process because the contact with oxygen is decreased when the slices are in water.

5.96 Zinc changing from zinc metal into zinc ions is an oxidation reaction because the zinc atoms are losing electrons to become cations. This reaction is the source of electrons as shown in the following equation:

Zn (s) → Zn2+ (aq) + 2 e-

EXAM QUESTIONS MULTIPLE CHOICE

1. Which of the following statements is true about the reaction?

2 CH4 + 3 O2 → 2 CO2 + 4 H2O

a. Water is a reactant.

b. Methane and carbon dioxide are products.

c. Methane is a product

d. Water and carbon dioxide are products

Answer: D

2. Which of the following reactions is correctly balanced?

a. N2 + H2 → 2NH3

Answer: C

110 Chapter 5

5.94 2 Cr2O7 2- (aq) + 3 CH3CH2OH (aq)

→

Zn + 2HCl

+ ZnCl

→

2H2O + C

CO + 2H2

+ O2 → CO2

d. CO

3. Which set of coefficients balances the equation?

4. When the equation below is properly balanced, what coefficient is in front of KCl?

5. Propane, C3H8, burns to produce carbon dioxide and water by the equation below. What is the coefficient in front of the carbon dioxide in the balanced equation?

6. Consider the reactants and propose the right side of the equation.

Chemical Reactions 111

 CH4 →  C3H8 +  H2 a. 3,1,1 b. 3,2,1 c. 3,1,2 d. 6,2,2 Answer:

KClO3 → KCl + O2 a. 1 b. 2 c. 3 d. 4 Answer:

C3H8 + O2 → CO2 + H2O a. 1 b. 2 c. 3 d. 6 Answer:

Mg + H2SO4 →  a. no reaction b. MgHSO4 c. MgSO4 + H2 d. MgH2 + SO4 Answer:

7. Consider the reactants in the partial equation given. Which choice is a product?

8. Which species would be considered a spectator ion based on the following equation?

9. A redox reaction can also be a:

a. combination reaction.

decomposition reaction. b. single replacement reaction

more than one response is correct

10. Which of the following is typically an oxidizing agent?

oxygen

hydrogen

water

hydrogen chloride

11. Single replacement reactions are always:

redox reactions.

nonredox reactions.

combination reactions.

decomposition reactions.

12. Which of the following equations (not balanced) represents an oxidation-reduction process?

13. What is the oxidation number of Cl in HClO3?

14. What is the oxidation number of Br in BrO ?

15. What is the oxidation number of Mn in KMnO4?

16. In which of the following does an element have an oxidation number of +4?

17. Which of the following contains the metal with the highest oxidation number?

112 Chapter 5

H3PO4 + Sr(OH)2 →  a. no reaction b. SrH3PO5 c. SrH5PO6 d. Sr3(PO4)2 Answer:

Ba2+ + 2NO3 - + 2Na+ + SO4 2+ → BaSO4(s) + 2NO3 - + 2Na+ a. Ba2+ b. SO4 2+ c. BaSO4 d. 2NO3Answer:

Answer:

a. AgNO3 + NaCl → AgCl + NaNO3 c. C2H4 + O2 → CO2 + H2O

C + O2 → CO2 d. more than

one response is correct Answer:

a. +2 b. +3 c. +4 d. +5 Answer:

a. -1 b. +1 c. +2 d. +3 Answer:

a. +9 b. +7 c. +5 d. +4 Answer:

TiO2 b. HBr c. SO3 d. SO

Answer:

a. CaCl2 b. NaCl c. FeCl3 d. CuCl2 Answer:

Reaction: The following questions are based on the following reaction.

18. Refer to Reaction. What substance is oxidized?

Answer: A

19. Refer to Reaction. What substance is reduced?

20. Refer to Reaction. What is the oxidizing agent?

21. Refer to Reaction. What is the reducing agent?

22. Which of the following is a decomposition reaction?

23. Which of the following is a combination reaction?

24. Which of the following is a single replacement reaction?

25. Which of the following is a double replacement reaction?

Answer: B

26. Barium sulfate is often ingested to aid in the diagnosis of GI tract disorders. The equation for its production is Ba2+ + SO → BaSO4(s). What type of equation is this?

a. a total ionic equation

equation

b. a full equation

a net ionic equation Answer: D

Chemical Reactions 113

4HCl + MnO2 → Cl2 + 2H2O + MnCl2

a. Cl in HCl b. Mn in MnO2 c. H in HCl d. O in MnO2

Answer: B

a. HCl b. Cl2 c. MnO2 d. MnCl2 Answer:

a. SO2 + O2 → 2SO3 b. 2C2H2 + 5O2 → 4CO2 + 2H2O c. 2H2O2 → 2H2O + O2 d. AgNO3 + NaCl → AgCl + NaNO3 Answer: C

a. SO2 + O2 → 2SO3 c. 2H2O2 → 2H2O + O2 b. 2C2H2 + 5O2 → 4CO2 + 2H2O d. AgNO3 + NaCl → AgCl + NaNO3 Answer:

a. CuO + H2 → Cu + H2O c. SO2 + H2O → H2SO3 b. HBr + KOH → H2O + KBr d. 2HI → I2 + H2 Answer:

a. CuO + H2 → Cu + H2O c. SO2 + H2O → H2SO3 b. HBr + KOH → H2O + KBr d. 2HI → I2 + H2

c. an oxidation-reduction

114 Chapter 5

27. Which ions in the following reaction would be classified as spectator ions?

Zn + 2H+ + 2Cl- → Zn2+ + 2Cl- + H2

a. Zn2+

b. H+

Answer: C

c. Cl-

d. more than one response is correct.

28. Which ion or ions will appear in the following (total ionic) reaction when it is written in net ionic form? Ba2+ + CO + 2H+ + SO → CO2 + H2O + BaSO4

a. Ba2+ and CO

b. H+ and SO

Answer: D

c. Ba2+, CO , and H+

d. Ba2+, CO , H+, and SO

29. For the reaction of HCl and NaOH, which of the following type of chemicals will appear in the net equation?

a. weak acid

Answer: D

b. weak base c. salt d. water

30. In the equation for an endothermic reaction, the word "heat" appears:

a. on the right side

Answer: B

b. on the left side

31. The energy involved in chemical reactions:

a. always appears as heat

b. always appears as light

Answer: D

c. on both sides d. on neither side

c. always appears as electricity

d. can take many forms

32. Given the equation 2CO + O22CO2, how many grams of CO2 form if 34.0 grams of CO are combined with excess oxygen?

a. 53.4

Answer: A

21.6

145

107

33. Oxygen can be prepared in the lab by heating potassium chlorate in the presence of a catalyst. The reaction is 2KClO3 → 2KCl + 3O2. How many moles of O2 could be obtained from one mole of KClO3? a. 1.0

Answer: B

34. The Haber Process is a method for the production of ammonia, an important chemical intermediate. The reaction for this process is: N2 + 3H2 → 2NH3. When the equation is correctly interpreted in terms of moles, how many moles of H2 will react with one mole of N2?

Answer: C

35. How many grams of N2 are required to completely react with 3.03 grams of H2 for the following balanced chemical equation? N2 + 3H2 → 2NH3 a. 1.00

Answer: C

6.00

c. 2.0 d. 3.0

1.5

a. 1

14.0

28.0

36. Refer to Haber Process. According to the mole interpretation, how many grams of NH3 could be produced by the complete reaction of 2.80 grams of N2 and excess H2?

a. 34.0 b. 17.0 c. 3.10 d. 3.40

Answer: D

37. In the reaction 2H2O → 2H2 + O2, 2.0 mol water will produce how many grams of O2?

a. 16 b. 32 c. 36 d. 64

Answer: B

38. In the reaction 2CH4 + O2 → 2CO + 4H2, how many molecules of CO will be produced by the complete reaction of 1.60 grams of CH4?

a. 12.0 x 1023 b. 3.01 x 1022

Answer: C

6.02 x 1022 d. 2.00

39. If 5 grams of CO and 5 grams of O2 are combined according to the reaction 2CO + O2 → 2CO2, which is the limiting reagent?

a. CO2 b. O2

Answer: C

40. Acetylene, C2H2, burns according to the following reaction: C

+5O

CO and O2

+2H

Suppose 1.20 g of C2H2 is mixed with 3.50 g of O2 in a closed, steel container, and the mixture is ignited. What substances will be found in the mixture when the burning is complete?

41. Suppose 2.43 g of magnesium is reacted with 3.20 g of oxygen: 2Mg + O2 → 2MgO. How many grams of MgO will be produced?

a. 5.63 b. 1.27

Answer: D

42. Tincture of iodine is an antiseptic. What is a tincture?

a. iodine dissolved in water

b. a substance dissolved in alcohol

Answer: B

a 3% hydrogen peroxide solution

an oxidizing antiseptic

43. What coefficients are needed to balance the following reaction: CH4 + ______ O2 → ______ CO2 + ______ H2O

a. 4,1,2 b. 2,0.5,1

Answer: C

none of these

44. According to the law of conservation of matter, which of the following may NEVER occur as a result of a chemical reaction:

a. creation of new chemical species.

b. reaction of starting materials

Answer: D

c. rearranged to form new substances.

d. reduction in total mass

Chemical Reactions 115

→ 4CO

CO2

H2 c. C2H2, CO2, and H2O

and

O2,

2, and H2O d. O2, C2H2, CO2, and H2O

Answer: C

c. 8.06 d.

4.03

2,1,2

116 Chapter 5

45. Which of the following is not a common use of the term oxidation?

a. to gain electrons

b. to lose hydrogen

Answer: A

46. The modern definition of oxidation is a process:

a. involving a reaction with oxygen

b. in which hydrogen is gained

Answer: D

c. to combine with oxygen

d. to increase in oxidation number

c. in which electrons are gained

d. none of the above

47. Tests are to be conducted on new anti-cancer drug and you are to prepare a fresh supply for the human subject studies. You discover that you were able to produce 1.345 g of this substance although it was theoretically possible to make 1.433 g. What was the percent yield for this synthesis?

a. 93.859 %

b. 93.86 %

Answer: B

c. 94 %

d. 0.9386 %

48. A percentage yield less than 100% is often obtained in a reaction. Which of the following could be the reason(s) for obtaining less than theoretically predicted?

a. side reactions occurred

b. poor laboratory technique

Answer: D

c. reaction did not go to completion

d. all of the above

49. A percentage yield greater than 100% has been reported in a scientific paper you are assigned to read. What would be a reasonable explanation for this?

a. A lack of training of the professor who published the study.

b. The product was measured more than once.

c. Small measurement errors could have caused this to occur.

d. It must be a typographical error since yields can’t exceed 100%

Answer: C

50. Which of the following types of chemical reaction classes always involve oxidation?

a. single replacement

b. double replacement

Answer: A

TRUE-FALSE

c. combination

d. decomposition

1. The reaction Mg + 3N2 → Mg3N2 is correctly balanced as written.

Answer: F

2. The reaction Ca + Cl2 → CaCl2 is a redox reaction.

Answer: T

3. Redox reactions may also be decomposition reactions

Answer: T

4. The oxidation number of iodine in HIO4 is +6.

Answer: F

5. In the reaction Zn + Cu(NO3)2 → Cu + Zn(NO3)2, the Zn is reduced.

Answer: F

6. In the reaction CuCl2 + Fe → Cu + FeCl2 the CuCl2 is the oxidizing agent.

Answer: T

7. The reaction AgNO3 + HCl → AgCl + HNO3 is a redox reaction.

Answer: F

8. In an exothermic reaction, heat is liberated to the surroundings.

Answer: T

9. All redox reactions are exothermic reactions.

Answer: F

10 The net ionic equation for the reaction

Answer: T

11. The oxidation number of oxygen in H2O2 is -2.

Answer: F

12. Products are always written on the right side of a chemical equation.

Answer: T

13 In ionic compounds, metals have a positive oxidation number.

Answer: T

14. The reducing agent is the substance that is reduced in the reaction.

Answer: F

15. Some redox reactions involve only oxidation without any reduction.

Answer: F

16. A net ionic equation may include one or more spectator ions.

Answer: F

17. The following reactions have the same net ionic equation.

NaOH (aq) + HCl (aq) → H2O (l) + NaCl (aq) and Ca(OH)2 (aq) + 2HBr (aq) → 2H2O (l) + CaBr2 (aq)

Answer: T

18. Most reactions give a percent yield of almost 100%.

Answer: F

19. Water is a product when any inorganic acid reacts with any inorganic base.

Answer: T

20. The batteries in an automobile contain lead plates, water, PbSO4 and H2SO4. The instructions with a battery charger state that when a car battery is being charged, there must be no flames or sparks. Hydrogen is liberated during the process and might explode on ignition. Therefore, the hydrogen released is an indication that a redox reaction occurs during the charging process.

Answer: T

Chemical Reactions 117

H2SO4 + 2 NaOH → Na2SO4 + 2 H2O is H+ + OHH2O.