Chapter 5: Chemical Reactions CHAPTER OUTLINE 5.1 5.2 5.3 5.4 Chemical Equations Types of Reactions Redox Reactions Decomposition Reactions 5.5 5.6 5.7 5.8 Combination Reactions Replacement Reactions Ionic Equations Energy and Reactions 5.9 The Mole and Chemical Equations 5.10 The Limiting Reactant 5.11 Reaction Yields LEARNING OBJECTIVES/ASSESSMENT 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 91 92 Chapter 5 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 5.1) Reactants 5.1 BaO2, H2SO4 a. BaO2 (s) + H2SO4 (l) → BaSO4 (s) + H2O2 (l) b. 2 H2O2 (aq) → 2 H2O (l) + O2 (g) H2O2 c. methane + water → carbon monoxide + hydrogen methane, water d. copper(II) oxide + hydrogen → copper + water copper(II) oxide, hydrogen ;5.2 Reactants a. H2 (g) + Cl2 (g) → 2 HCl (g) H2, Cl2 b. 2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) KClO3 c. magnesium oxide + carbon → magnesium magnesium + carbon monoxide oxide, carbon d. ethane + oxygen → carbon dioxide + water ethane, oxygen Products BaSO4, H2O2 H2O, O2 carbon monoxide, hydrogen copper, water Products HCl KCl, O2 magnesium, carbon monoxide 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. ⎛ 1 mole O 2 ⎞
Re ac tan ts : 3.20 g O2 ⎜⎜
⎟⎟ = 0.100 moles O 2 ; ⎝ 32.0 g O 2 ⎠
⎛ 1 mole S ⎞
3.21 g S ⎜⎜
⎟⎟ = 0.100 moles S
⎝ 32.1 g S ⎠
⎛ 1 mole SO 2 ⎞⎛ 2 moles O ⎞
Products : 6.41 g SO 2 ⎜⎜
⎟⎜
⎟⎟ = 0.200 moles O; ⎟⎜
⎝ 64.1 g SO2 ⎠⎝ 1 mole SO 2 ⎠
⎛ 1 mole SO 2 ⎞ ⎛ 1 mole S ⎞
6.41 g SO 2 ⎜⎜
⎟⎟ ⎜⎜
⎟⎟ = 0.100 moles S
⎝ 64.1 g SO 2 ⎠ ⎝ 1 mole SO 2 ⎠
d. CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) is consistent with the law of conservation of matter. 5.4 a. 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. Chemical Reactions 93 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. ⎛ 1 mole O2 ⎞
Re ac tan ts : 1.50 g O 2 ⎜⎜
⎟⎟ = 0.0469 moles O 2 ; ⎝ 32.0 g O 2 ⎠
⎛ 1 mole C ⎞
1.50 g C ⎜⎜
⎟⎟ = 0.125 moles C
⎝ 12.0 g C ⎠
⎛ 1 mole CO ⎞ ⎛ 1 mole O ⎞
Products : 2.80 g CO ⎜⎜
⎟⎟ ⎜⎜
⎟⎟ = 0.100 mole O;
⎝ 28.0 g CO ⎠ ⎝ 1 mole CO ⎠
⎛ 1 mole CO ⎞ ⎛ 1 mole C ⎞
2.80 g CO ⎜⎜
⎟⎟ ⎜⎜
⎟⎟ = 0.100 mole C
⎝ 28.0 g CO ⎠ ⎝ 1 mole CO ⎠
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 1 Product 1 2 This equation is balanced. c. SO2 (g) + H2O (l) → H2SO3 (aq) Elements S O H Reactant 1 3 2 Product 1 3 2 This equation is not balanced because the numbers of moles of oxygen atoms are not balanced. This equation is balanced. d. Ba(ClO3)2 (aq) + H2SO4 (aq) → 2HClO3 (aq) + BaSO4 (s) Elements Ba Cl O H S Reactant 1 2 10 2 1 This equation is balanced. Product 1 2 10 2 1 ;5.6 a. 2 Ag2O (s) → 2 Ag (s) + O2 (g) Elements Ag O This equation is not balanced because the numbers of moles of Reactant 4 2 silver atoms are not balanced. Product 2 2 b. Al (s) + O2 (g) → Al2O3 (s)
Elements Al O Reactant 1 2 Product 2 3 This equation is not balanced because the numbers of moles of aluminum and oxygen atoms are not balanced. 94 Chapter 5 c. 2 AgNO3 (aq) + K2SO4 (aq) → Ag2SO4 (s) + 2 KNO3 (aq)
Elements Ag N O K S Reactant 2 2 10 2 1 This equation is balanced. Product 2 2 10 2 1 d. SO2 (g) + O2 (g) → SO3 (g) Elements S O This equation is not balanced because the numbers of moles of Reactant 1 4 oxygen atoms are not balanced. Product 1 3 5.7 a. Cl2 (aq) + NaBr (aq)→ NaCl (aq) + Br2 (aq) b. CaF2 (s) + H2SO4 (aq) → CaSO4 (s) + HF (g) c. Cl2 (g) + NaOH (aq) → NaOCl (aq) + NaCl (aq) + H2O (l) d. KClO3 (s) → KClO4 (s) + KCl (s) e. dinitrogen monoxide → nitrogen + oxygen f. dinitrogen pentoxide → nitrogen dioxide + oxygen g. P4O10 (s) + H2O (l) → H4P2O7 (aq) h. CaCO3 (s) + HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g) Cl2 (aq) + 2 NaBr (aq)→ 2 NaCl (aq) + Br2 (aq) CaF2 (s) + H2SO4 (aq) → CaSO4 (s) + 2 HF (g) Cl2 (g) + 2 NaOH (aq) → NaOCl (aq) + NaCl (aq) + H2O (l) 4 KClO3 (s) → 3 KClO4 (s) + KCl (s) 2 N2O → 2 N2 + O2 2 N2O5 → 4 NO2 + O2 P4O10 (s) + 4 H2O (l) → 2 H4P2O7 (aq) CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g) 5.8 a. b. c. d. e. C2H6 (g) + O2 (g) → CO2(g) + H2O (l) 2 C2H6 (g) + 7 O2 (g) → 4 CO2(g) + 6 H2O (l) hydrogen + chlorine → hydrogen chloride H2 (g) + Cl2 (g) → 2 HCl (g) H2S (g) + O2 (g) → SO2 (g) + H2O (l) 2 H2S (g) + 3 O2 (g) → 2 SO2 (g) + 2 H2O (l) sulfur + oxygen → sulfur dioxide S (s) + O2 (g) → SO2 (g) Na2CO3 (aq) + Ca(NO3)2 (aq) → Na2CO3 (aq) + Ca(NO3)2 (aq) → NaNO3 (aq) + CaCO3 (s) 2 NaNO3 (aq) + CaCO3 (s) f. NaBr (aq) + Cl2 (aq) → NaCl (aq) + Br2 (aq) 2NaBr (aq) + Cl2 (aq) → 2NaCl (aq) + Br2 (aq) 2 Ag2CO3 (s) → 4 Ag (s) + 2 CO2 (g) + O2 (g) g. Ag2CO3 (s) → Ag (s) + CO2 (g) + O2 (g) h. H2O2 (aq) + H2S (aq) → H2O (l) + S (s) H2O2 (aq) + H2S (aq) → 2 H2O (l) + S (s) REDOX REACTIONS (SECTION 5.3) 5.9 Cl = ?
a. Cl2O5 O = −2
2 ( ? ) + 5( −2) = 0
Calculations Cl 2 O5 = 0
⇒ ? = +5
+5 b. KClO4 K = + 1 Cl = ?
+1 + ?+ 4( −2) = 0
c. Ba2+ d. F2 e. H4P2O7 +2 H = +1 P = ? O = −2 H 4 P2 O7 = 0
0 O = −2
KClO4 = 0
⇒ ? = +7
4 ( +1) + 2 ( ? ) + 7 ( −2 ) = 0
f. H2S H = +1
S=?
2 ( +1) + ? = 0
O.N. ⇒ ? = +5
H2 S = 0
⇒ ? = −2
+7 +5 ‐2 Chemical Reactions 95 ;5.10 a. ClO3‐ b. H2SO4 Calculations ClO 3 − = −1
?+ 3( −2) = −1 ⇒ ? = + 5
H = +1 S = ?
O = −2
H 2 SO 4 = 0
Cl = ?
2 ( +1) + ?+ 4( −2) = 0
c. NaNO3 ⇒ ? = +6
Na = +1 N = ? O = −2 NaNO 3 = 0
( +1) + ?+ 3 ( −2 ) = 0 ⇒ ? = +5
d. N2O N=?
O = −2
2 ( ? ) + ( −2) = 0
e. KMnO4 N2 O = 0
⇒ ? = +1
f. HClO2 H = +1
Cl = ?
( +1) + ?+ 2( −2) = 0
+5 +6 +5 +1 K = +1 Mn = ? O = −2 KMnO 4 = 0
( +1) + ?+ 4 ( −2 ) = 0
O.N. O = −2
⇒ ? = +7
O = −2
HClO2 = 0
⇒ ?=+3
+7 +3 Oxidation Numbers N = +5, O = ‐2 K = +1, H = +1, C = +4, O = ‐2 Na = +1, O = ‐2, Cl = +1 Na = +1, N = +5, O = ‐2 H = +1, Cl = +7, O = ‐2 Ca = +2, N = +5, O = ‐2 Highest O.N. nitrogen carbon sodium and chlorine nitrogen chlorine nitrogen Oxidation Numbers Na = +1, Cr = +6, O = ‐2 K = +1, S = +2, O = ‐2 H = +1, N = +5, O = ‐2 P = +5, O = ‐2 Mg = +2, Cl = +7, O = ‐2 H = +1, Cl = +3, O = ‐2 Highest O.N. chromium sulfur nitrogen phosphorus chlorine chlorine 5.11 a. b. c. d. e. f. 5.12 a. b. c. d. e. f. 5.13 a. b. c. d. e. 2 Mg (s) + O2 (g) → 2 MgO (s) CuO (s) + H2 (g) → Cu (s) + H2O (g) Ag+ (aq) + Cl‐ (aq) → AgCl (s) BaCl2(aq) + H2SO4 (aq) → BaSO4 (s) + 2 HCl (aq) Zn (s) + 2 H+ (aq) → Zn2+ (aq) + H2 (g) 5.14 a. b. c. d. e. 4 Al (s) + 3 O2 (aq) → 2 Al2O3 (s) SO2 (g) + H2O (l) → H2SO4 (aq) 2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) 2 CO (g) + O2 (g) → 2 CO2 (g) 2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g) N2O5 KHCO3 NaOCl NaNO3 HClO4 Ca(NO3)2 Na2Cr2O7 K2S2O3 HNO3 P2O5 Mg(ClO4)2 HClO2 O.N. 0→+2 +2→0 +1→+1 +2→+2 0→+2 Change Classification lost e‐ oxidized gained e‐ reduced no change neither no change neither lost e‐ oxidized O.N. 0 → +3 +1 → +1 +5 → ‐1 +2 → +4 0 → +1 Change Classification lost e‐ oxidized no change neither gained e‐ reduced lost e‐ oxidized lost e‐ oxidized 96 Chapter 5 ;5.15 a. H2 (g) + Cl2 (g) → 2 HCl (g) Elements H Cl Reactant 0 0 Product +1 ‐1 The oxidizing agent is Cl2. The reducing agent is H2. b. H2O (g) + CH4 (g) → CO (g) + 3 H2 (g) Elements H O C The oxidizing agent is H2O. Reactant +1 ‐2 ‐4 The reducing agent is CH4. 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 The oxidizing agent is CuO. The reducing agent is H2. d. B2O3 (s) + 3 Mg (s) → 2 B (s) + 3 MgO (s) Elements B O Mg The oxidizing agent is B2O3. Reactant +3 ‐2 0 The reducing agent is Mg. 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 Fe2O3. The reducing agent is CO. f. Cr2O72‐ (aq) + 2 H+ (aq) + 3 Mn2+ (aq) → 2 Cr3+ (aq) + 3 MnO2 (s) + H2O (l) Elements Cr O H Mn 2‐
The oxidizing agent is Cr2O7 . Reactant +6 ‐2 +1 +2 2+
The reducing agent is Mn . 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 O2. The reducing agent is Cu. b. Cl2 (aq) + 2 KI (aq) → 2 KCl (aq) + I2 (aq) Elements Cl K I The oxidizing agent is Cl2. Reactant 0 +1 ‐1 The reducing agent is KI. Product ‐1 +1 0 c. 3 MnO2 (s) + 4 Al (s) → 2 Al2O3 (s) + 3 Mn (s) Elements Mn O Al The oxidizing agent is MnO2. Reactant +4 ‐2 0 The reducing agent is Al. Product 0 ‐2 +3 Chemical Reactions 97 d. 2 H+ (aq) + 3 SO32‐ (aq) + 2 NO3‐ (aq) → 2 NO (g) + H2O (l)+ 3 SO42‐ (aq) Elements H S O N ‐
The oxidizing agent is NO3 . Reactant +1 +4 ‐2 +5 2‐
The reducing agent is SO3 . Product +1 +6 ‐2 +2 e. Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g) Elements Mg H Cl The oxidizing agent is HCl. Reactant 0 +1 ‐1 The reducing agent is Mg. Product +2 0 ‐1 f. 4 NO2 (g) + O2 (g) → 2 N2O5 (g) Elements N O O Reactant +4 ‐2 0 Product +5 ‐2 ‐2 The oxidizing agent is O2. The reducing agent is NO2. 5.17 3 Ag2S (s) + 2 Al (s) → 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) ;5.20 a. K2CO3 (s) → K2O (s) + CO2 (g) nonredox: decomposition b. Ca (s) + 2 H2O (l) → Ca(OH)2 (s) + H2 (g) redox: single‐replacement 0 +1 ‐2 +2 ‐2 +1 0 c. BaCl2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2 HCl (aq) d. SO2 (g) + H2O (l) →H2SO3 (aq) e. 2 NO (g) + O2 (g) → NO2 (g) +2 ‐2 0 +4 ‐2 f. 2 Zn (s) + O2 (g) → 2 ZnO (s) 0 0 +2 ‐2 nonredox: double‐replacement nonredox: combination redox: combination a. N2O5 (g) + H2O (l) → 2 HNO3 (aq) b. Cr2O3 (s) + 2 Al (s) → 2 Cr(s) + Al2O3 (s) +3 ‐2 0 0 +3 ‐2 c. CaO (s) + SiO2 (s) → CaSiO3 (s) d. H2CO3 (aq) → CO2 (g) + H2O (l) e. PbCO3 (s) → PbO (s) + CO2 (g) f. Zn (s) + Cl2 (g) → ZnCl2 (s) 0 0 +2 ‐1 nonredox: combination redox: single‐replacement redox: combination 5.21 nonredox: nonredox: nonredox: redox: combination decomposition decomposition combination 98 Chapter 5 Heat
2 NaHCO3 (s) ⎯⎯⎯
→ Na 2 CO3 (s) + H 2 O (g) + CO2 (g) 5.22 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. 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); This reaction is a nonredox combination reaction. IONIC EQUATIONS (SECTION 5.7) +
‐
5.28 a. LiNO3 Li , NO3 b. Na2HPO4 2 Na , HPO4 c. Ca(ClO3)2 Ca , 2 ClO3 5.29 a. K2Cr2O7 2 K , Cr2O7 b. H2SO4 2 H , SO4 c. NaH2PO4 Na , H2PO4 5.30 ; a. Cl2 (aq) + 2NaI (aq) → 2NaCl (aq) + I2 (aq) +
2‐
2+
‐
+
2‐
+
‐
d. KOH K , OH e. MgBr2 Mg , 2 Br f. (NH4)2SO4 2 NH4 , SO4 d. Na3PO4 3 Na , PO4 e. NH4Cl NH4 , Cl f. KMnO4 K , MnO4 2+
‐
+
2‐
+
2‐
+
‐
3‐
+
+
+
‐
‐
Total ionic equation: Cl2 (aq) + 2Na+ (aq) + 2 I‐ (aq) → 2Na+ (aq) + 2 Cl‐ (aq) + I2 (aq) Spectator ions: 2Na+ (aq) Net ionic equation: Cl2 (aq) + 2 I‐ (aq) → 2 Cl‐ (aq) + I2 (aq) ; b. AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq) Total ionic equation: ‐
‐
Ag+ (aq) + NO3 (aq) + Na+ (aq) + Cl‐ (aq) → AgCl (s) + Na+ (aq) + NO3 (aq) Spectator ions: NO3‐ (aq), Na+ (aq) Net ionic equation: Ag+ (aq) + Cl‐ (aq) → AgCl (s) ; c. Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2(g) Total ionic equation: Zn (s) + 2 H+ (aq) + 2 Cl‐ (aq) → Zn2+ (aq) + 2 Cl‐ (aq) + H2(g) Spectator ions: Cl‐ (aq) Net ionic equation: Zn (s) + 2 H+ (aq) → Zn2+ (aq) + H2(g) Chemical Reactions 99 d. BaCl2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2 HCl (aq) Total ionic equation: Ba2+(aq) + 2Cl‐(aq) + 2H+(aq) + SO42‐(aq) → BaSO4(s) + 2H+(aq) + 2Cl‐(aq) Spectator ions: 2Cl‐(aq) + 2H+(aq) Net ionic equation: Ba2+(aq) + SO42‐(aq) → BaSO4(s) e. SO3 (aq) + H2O (l) → 2 H2SO4 (aq) Total ionic equation: SO3 (aq) + H2O (l) → 4 H+ (aq) + 2 SO42‐ (aq) Spectator ions: none Net ionic equation: SO3 (aq) + H2O (l) → 4 H+ (aq) + 2 SO42‐ (aq) f. 2 NaI (aq) + 2 H2SO4 (aq)→ I2 (aq) + SO2 (aq) + Na2SO4 (aq) + 2 H2O (l) Total ionic equation: 2‐
2‐
2 Na+ (aq) + 2 I‐ (aq) + 4 H+ (aq) + 2 SO4 (aq)→ I2 (aq) + SO2 (aq) + 2 Na+ (aq) + SO4 (aq) + 2 H2O (l) 2‐
Spectator ions: 2Na+ (aq), SO4 (aq) Net ionic equation: 2‐
2 I‐ (aq) + 4 H+ (aq) + SO4 (aq)→ I2 (aq) + SO2 (aq) + 2 H2O (l) 5.31 a. H2O (l) + Na2SO3 (aq) + SO2 (aq) → 2NaHSO3 (aq) Total ionic equation: H2O (l) + 2Na+(aq) + SO32‐(aq) + SO2(aq) → 2Na+(aq) + 2 HSO3‐ (aq) Spectator ions: 2Na+ (aq) Net ionic equation: H2O (l) + SO32‐ (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) 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) 100 Chapter 5 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‐ +
+
+
‐
3‐ 3 H (aq) + PO4 (aq) + 3 NH4 (aq) + 3 OH (aq) → 3 NH4 (aq) + PO4 (aq) + 3 H2O (l) Spectator ions: 3 NH4+ (aq), PO43‐ (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) The reactants and products are identical for all three net ionic equations. 5.33 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) Spectator ions: Br‐(aq), Rb+ (aq) Net ionic equation: H+ (aq) + OH‐(aq) → H2O (l) b. H2SO4 (aq) + 2 LiOH (aq) → Li2SO4 (aq) + 2 H2O (l) Total ionic equation: +
2‐ +
‐
+
2‐ 2 H (aq) + SO4 (aq) + 2 Li (aq) + 2 OH (aq) → 2 Li (aq) + SO4 (aq) + 2 H2O (l) Spectator ions: SO42‐ (aq), 2 Li+ (aq) Net ionic equation: 2 H+ (aq) + 2 OH‐(aq) → 2 H2O (l); H+ (aq) + OH‐(aq) → H2O (l) Chemical Reactions 101 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) 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. THE MOLE AND CHEMICAL EQUATIONS (SECTION 5.9) 5.38 a. Ca (s) + 2 H2O (l) → H2 (g) + Ca(OH)2 (s) Statement 1: 1 Ca atom + 2 H 2 O molecules → 1 H 2 molecule + 1 Ca(OH)2 formula unit
Statement 2: 1 mole Ca + 2 moles H 2 O → 1 mole H 2 + 1 mole Ca(OH)2
Statement 3: 6.02 × 10 23 Ca atoms + 1.20 × 10 24 H 2 O molecules → 6.02 × 10 H 2 molecules + 6.02 × 10 Ca(OH)2 formula units
Statement 4 : 40.1 g Ca + 36.0 g H2 O → 2.02 g H 2 + 74.1 g Ca(OH)2
23
23
b. 2 NO (g) + O2 (g) → 2 NO2 (g) Statement 1: 2 NO molecules + 1 O2 molecules → 2 NO2 molecule
Statement 2: 2 moles NO + 1 mole O2 → 2 moles NO2
Statement 3: 1.20 × 1024 NO molecules + 6.02 × 1023 O2 molecules → 1.20 × 10 24 NO 2 molecules
Statement 4 : 60.0 g NO + 32.0 g O 2 → 92.0 g NO 2
c. 2 C2H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (l) Statement 1: 2 C2 H6 molecules + 7 O 2 molecules → 4 CO 2 molecules + 6 H 2 O molecules
Statement 2: 2 moles C 2 H6 + 7 moles O 2 → 4 moles CO 2 + 6 moles H 2 O
Statement 3: 1.20 × 10 24 C 2 H6 molecules + 4.22 × 10 24 O 2 molecules → 2.41 × 10
24
CO 2 molecules + 3.61 × 10
Statement 4 : 60.1 g C2 H6 + 224 g O 2 → 176 g CO 2 + 108 g H 2 O
24
H 2 O molecules
102 Chapter 5 d. Zn (s) + 2 AgNO3 (g) → Zn(NO3)2 (aq) + 2 Ag (s) Statement 1: 1 Zn atom + 2 AgNO 3 formula units → 1 Zn(NO 3 )2 formula unit + 2 Ag atoms
Statement 2: 1 mole Zn + 2 moles AgNO 3 → 1 mole Zn(NO 3 )2 + 2 moles Ag
Statement 3: 6.02 × 10 23 Zn atoms + 1.20 × 10 24 AgNO 3 formula units → 6.02 × 10
23
Zn(NO3 )2 formula unit + 1.20 × 10
24
Ag atoms
Statement 4 : 65.4 g Zn + 340 g AgNO 3 → 189 g Zn(NO3 )2 + 216 g Ag
e. 2 HCl (aq) + Mg(OH)2 (s) → MgCl2 (aq) + 2 H2O (l) Statement 1: 2 HCl molecules + 1 Mg(OH)2 formula unit → 1 MgCl 2 formula unit + 2 H 2 O molecules
Statement 2: 2 moles HCl + 1 mole Mg(OH)2 → 1 mole MgCl 2 + 2 moles H 2 O
Statement 3: 1.20 × 10 24 HCl molecules + 6.02 × 10 23 Mg(OH)2 formula units → 6.02 × 10 23 MgCl 2 formula units + 1.20 × 10 24 H 2 O molecules
Statement 4 : 73.0 g HCl + 58.3 g Mg(OH)2 → 95.3 g MgCl 2 + 36.0 g H 2 O
5.39 a. N2 (g) + 3 H2 (g) → 2NH3 (g) Statement 1: 1 N 2 molecule + 3 H 2 molecules → 2 NH 3 molecules Statement 2: 1 mole N 2 + 3 moles H 2 → 2 moles NH 3
Statement 3: 6.02 × 10 23 N 2 molecules + 1.81 × 10 24 H 2 molecules → 1.20 × 10 24 NH3 molecules Statement 4 : 28.0 g N 2 + 6.06 g H 2 → 34.1 g NH 3 b. 2 Na (s) + Cl2 (g) → 2 NaCl (s) Statement 1: 2 Na atoms + 1 Cl 2 molecule → 2 NaCl formula units
Statement 2: 2 moles Na + 1 mole Cl 2 → 2 moles NaCl
Statement 3: 1.20 × 10 24 Na atoms + 6.02 × 10 23 Cl 2 molecules → 1.20 × 10 24 NaCl formula units
Statement 4 : 46.0 g Na + 71.0 g Cl 2 → 117 g NaCl
c. BaCl2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2 HCl (aq) Statement 1: 1 BaCl 2 formula unit + 1 H 2 SO 4 molecule → 1 BaSO 4 formula unit + 2 HCl molecules
Statement 2: 1 mole BaCl 2 + 1 mole H 2 SO 4 → 1 mole BaSO 4 + 2 moles HCl
Statement 3: 6.02 × 10 23 BaCl 2 formula units + 6.02 × 10 23 H 2 SO 4 molecules → 6.02 × 10 23 BaSO 4 formula units + 1.20 × 10 24 HCl molecules
Statement 4 : 208 g BaCl 2 + 98.1 g H 2 SO 4 → 233 g BaSO 4 + 73.0 g HCl
d. 2 H2O2 (aq) → 2 H2O (l) + O2 (g) Statement 1: 2 H 2 O 2 molecules → 2 H 2 O molecules + 1 O2 molecule
Statement 2: 2 moles H 2 O 2 → 2 moles H 2 O + 1 mole O 2
Statement 3: 1.20 × 10 24 H 2 O 2 molecules → 12.0 × 10
24
H 2 O molecules + 6.02 × 10 O2 molecules
Statement 4 : 68.0 g H 2 O 2 → 36.0 g H 2 O + 32.0 g O 2
23
Chemical Reactions 103 e. 2 C3H6 (g) + 9 O2 (g) → 6 CO2 (g) + 6 H2O (g) Statement 1: 2 C 3 H6 molecules + 9 O 2 molecules → 6 CO 2 molecules + 6 H 2 O molecules
Statement 2: 2 moles C 3 H6 + 9 moles O 2 → 6 moles CO2 + 6 moles H 2 O
Statement 3: 1.20 × 10 24 C 3 H6 molecules + 5.42 × 10 24 O2 molecules → 3.61 × 10 24 CO 2 molecules + 3.61 × 10 24 H 2 O molecules
Statement 4 : 84.1 g C 3 H6 + 288 g O 2 → 264 g CO 2 + 108 g H 2 O
5.40 2 SO2 + O2 → 2 SO3 (g) Statement 1: 2 SO 2 molecules + 1 O 2 molecule → 2 SO 3 molecules
Statement 2: 2 moles SO2 + 1 mole O2 → 2 moles SO3
Statement 3: 12.0 × 10 23 SO 2 molecules + 6.02 × 10 23 O2 molecules → 12.0 × 10 SO3 molecules
23
Statement 4 : 128 g SO2 + 32.0 g O 2 → 160 g SO 3
Factors :
12.0 × 10 23 SO 2 molecules 6.02 × 10 23 O2 molecules 12.0 × 10 23 SO 2 molecules
;
;
;
6.02 × 10 23 O 2 molecules 12.0 × 10 23 SO 2 molecules 12.0 × 10 23 SO3 molecules
12.0 × 10 23 SO 3 molecules 6.02 × 10 23 O 2 molecules 12.0 × 10 23 SO 3 molecules
;
;
;
12.0 × 10 23 SO 2 molecules 12.0 × 10 23 SO 3 molecules 6.02 × 10 23 O 2 molecules
2 moles SO 2 1 mole O 2 2 moles SO 2 2 moles SO 3 1 mole O 2 2 moles SO 3
;
;
;
;
;
;
1 mole O 2 2 moles SO 2 2 moles SO 3 2 moles SO 2 2 moles SO 3 1 mole O 2
128 g SO 2 32.0 g O 2 128 g SO 2 160 g SO 3 32.0 g O 2 160 g SO 3
;
;
;
;
;
32.0 g O 2 128 g SO 2 160 g SO 3 128 g SO 2 160 g SO 3 32.0 g O 2
This list does not include all possible factors. 5.41 2 SO2 + O2 → 2 SO3 (g) Statement 1: 2 SO 2 molecules + 1 O 2 molecule → 2 SO 3 molecules
Statement 2: 2 moles SO2 + 1 mole O2 → 2 moles SO3
Statement 3: 12.0 × 10 23 SO 2 molecules + 6.02 × 10 23 O2 molecules → 12.0 × 10 SO3 molecules
23
Statement 4 : 128 g SO2 + 32.0 g O 2 → 160 g SO 3
⎛ 128 g SO 2
350 g SO 3 ⎜⎜
⎝ 160 g SO 3
⎞
⎟⎟ = 280 g SO 2
⎠
280 g SO2 must react to produce 350 g SO3. ;5.42 CaCO3 (s) → CaO (s) + CO2 (g) ⎛ 100 g CaCO 3 ⎞
500 g CaO ⎜⎜
⎟⎟ = 891.265597148 g CaCO 3
⎝ 56.1 g CaO ⎠
≈ 891 g CaCO3
104 Chapter 5 5.43 CaCO3 (s) → CaO (s) + CO2 (g) ⎛ 1 mole CO 2 ⎞
500 g CaO ⎜⎜
⎟⎟ = 8.91265597148 moles CO 2 ⎝ 56.1 g CaO ⎠
≈ 8.91 moles CO 2
5.44 2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s) ⎛ 479 g Br2 ⎞
50.1 g Al ⎜⎜
⎟⎟ = 444.4055556 g Br2 ⎝ 54.0 g Al ⎠
≈ 444 g Br2 with SF
5.45 2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s) ⎛ 2 moles AlBr3 ⎞
50.1 g Al ⎜⎜
⎟⎟ = 1.85555556 moles AlBr3 ⎝ 54.0 g Al ⎠
≈ 1.86 moles AlBr3 with SF
5.46 ⎛ 533 g AlBr3
50.1 g Al ⎜⎜
⎝ 54.0 g Al
⎞
⎟⎟ = 494.50555556 g AlBr3
⎠
≈ 495 g AlBr3 with SF
or
50.1 g Al + 445 g Br2 = 495.1 g Al Br3 ≈ 495 g AlBr3 (SF)
5.47 a. ⎛ 54.0 g Al ⎞
0.250 g Ag 2 S ⎜⎜
⎟⎟ = 0.01814516129 g Al
⎝ 744 g Ag 2 S ⎠
≈ 1.81 × 10 −2 g Al with SF
3 Ag2S (s) + 2 Al (s) → 6 Ag (s) + Al2S3 (s) b. ⎛ 1 mole Al 2 S 3
0.250 g Ag 2 S ⎜⎜
⎝ 744 g Ag 2 S
⎞
−4
⎟⎟ = 3.36021505 × 10 moles Al 2 S 3
⎠
−4
≈ 3.36 × 10 moles Al 2 S 3 with SF
5.48 TiCl4 (s) + 2 Mg (s) → Ti (s) + 2 MgCl2 (s) ⎛ 1000 g Ti ⎞⎛ 48.6 g Mg ⎞
1.00 kg Ti ⎜⎜
⎟⎜
⎟⎟ = 1014.61377871 g Mg
⎟⎜
⎝ 1 kg Ti ⎠⎝ 47.9 g Ti ⎠
3
≈ 1.01 × 10 g Mg with SF
5.49 C6H12O6 (aq) + 6 O2 (aq) → 6 CO2 (aq) + 6 H2O (l) a. ⎛ 108 g H 2 O ⎞
1.00 mole glu cos e ⎜⎜
⎟⎟ = 108 g H 2 O
⎝ 1 mole glu cos e ⎠
2
= 1.08 × 10 g H 2 O
b. ⎛
⎞
192 g O 2
1.00 mole glu cos e ⎜⎜
⎟⎟ = 192 g O2
⎝ 1 mole glu cos e ⎠
2
= 1.92 × 10 g O 2
Chemical Reactions 105 5.50 C6H12O2 (aq) + 8 O2 (aq) → 6 CO2 (aq) + 6 H2O (l) ⎛
⎞
256 g O2
1.00 mol caprylic acid ⎜⎜
⎟⎟ = 256 g O 2 ⎝ 1 mol caprylic acid ⎠
THE LIMITING REACTANT (SECTION 5.10) 5.51 a. 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: ⎛ 154 g CCl 4 ⎞
4.00 g CH 4 ⎜⎜
⎟⎟ = 38.5 g CCl 4
⎝ 16.0 g CH 4 ⎠
⎛ 154 g CCl 4 ⎞
15.0 g Cl 2 ⎜⎜
⎟⎟ = 8.13 g CCl 4
⎝ 284 g Cl 2 ⎠
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. ;5.52 a. N2 (g) + 2 O2 (g) → 2 NO2 (g) O2 is the limiting reactant as shown by the calculations below: ⎛ 2 moles NO 2 ⎞
1.25 moles N 2 ⎜⎜
⎟⎟ = 2.50 moles NO 2
⎝ 1 mole N 2 ⎠
⎛ 2 moles NO 2 ⎞
50.0 g O 2 ⎜⎜
⎟⎟ = 1.56 moles NO 2
⎝ 64.00 g O 2 ⎠
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 of N2. ⎛ 92.0 g NO 2 ⎞
50.0 g O 2 ⎜⎜
⎟⎟ = 71.9 g NO 2
⎝ 64.0 g O2 ⎠
⎛ 92.0 g NO 2 ⎞
1.25 moles N 2 ⎜⎜
⎟⎟ = 115 g NO 2
⎝ 1 mole N 2 ⎠
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. ⎛ 160 g O2 ⎞
3
500 g C2 H 2 ⎜⎜
⎟⎟ = 1538 g O 2 ≈ 1.54 × 10 g O 2 52.0
g
C
H
2
2 ⎠
⎝
5.54 NH3 (aq) + CO2 (aq) + H2O (l) → NH4HCO3 (aq) 144 g NH4HCO3 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 NH3 and H2O are both excess reactants. 106 Chapter 5 ⎛ 79.1 g NH 4 HCO 3 ⎞
50.0 g NH 3 ⎜⎜
⎟⎟ = 233 g NH 4 HCO 3
17.0 g NH 3
⎝
⎠
⎛ 79.1 g NH 4 HCO3 ⎞
80.0 g CO 2 ⎜⎜
⎟⎟ = 144 g NH 4 HCO 3
44.0 g CO2
⎝
⎠
⎛ 79.1 g NH 4 HCO3 ⎞
2.00 moles H 2 O ⎜⎜
⎟⎟ = 158 g NH 4 HCO 3
⎝ 1 moles H 2 O ⎠
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 Al. ⎛ 54.0 g Al ⎞
150 g Cr2 O3 ⎜⎜
⎟⎟ = 53.289 g Al ≈ 53.3 g Al ⎝ 152 g Cr2 O3 ⎠
REACTION YIELDS (SECTION 5.11) ;5.56 12.18 g
× 100 = 76.46% yield 15.93 g
5.57 14.37 g
× 100 = 81.88% yield 17.55 g
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. 9.04 g
× 100 = 72.5% yield 7.59 g + 4.88 g
5.59 2 Ca (s) + O2 (g) → 2 CaO (s) ⎛ 112 g CaO ⎞
2.00 g Ca ⎜⎜
⎟⎟ = 2.7930 g CaO
2.26 g
⎝ 80.2 g Ca ⎠
or × 100 = 80.9% yield ⎛ 112 g CaO ⎞
2.26 g
2.00 g Ca ⎜
× 100 = 80.9% yield
⎟
2.7930 g CaO
⎝ 80.2 g Ca ⎠
5.60 2 HgO (s) → 2 Hg (l) + O2 (g) ⎛ 402 g Hg ⎞
7.22 g HgO ⎜⎜
⎟⎟ = 6.6876 g Hg
5.95 g
⎝ 434 g HgO ⎠
× 100 = 89.0% yield or ⎛ 402 g Hg ⎞
5.95 g
7.22 g HgO ⎜
× 100 = 89.0% yield
⎟
6.6876 g Hg
⎝ 434 g HgO ⎠
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. Chemical Reactions 107 5.62 barium chloride (aq) + sodium sulfate (aq) → sodium chloride (aq) + barium sulfate (s) Molecular equation: BaCl2 (aq) + Na2SO4 (aq) → 2 NaCl (aq) + BaSO4 (s) Total Ionic equation: Ba 2 + (aq) + 2 Cl − (aq) + 2 Na + (aq) + SO42 − (aq) → 2 Na + (aq) + 2 Cl − (aq) + BaSO4 (s) Spectator Ions: 2 Na+ (aq) and 2 Cl‐ (aq) Net Ionic equation: Ba 2 + (aq) + SO24 − (aq) → BaSO 4 (s) 5.63 Sodium is the element with an electron configuration of 1s22s22p63s1. Phosphorus is the element with 15 protons in its nucleus. 3Na (s) + P (s) → Na 3 P(s) 5.64 ⎛
⎞
⎟
60
naturally occuring ⎜
60
g
Fe
⎜
⎟ = 7.5 g
6.983 g
naturally occuring ⎟
elemental iron ⎜
⎜ 55.9 g elemental iron ⎟
⎝
⎠
60
Fe 5.65 The formula of the sample that was decomposed was N2O5, dinitrogen pentoxide. ⎛ 1 mole O ⎞
80.0 g O ⎜⎜
⎟⎟ = 5.00 moles O
⎝ 16.0 g O ⎠
⎛
⎞
1 mole N
24
1.20 × 10 atoms N ⎜⎜
⎟⎟ = 1.99 moles N
23
⎝ 6.02 × 10 atoms N ⎠
ALLIED HEALTH EXAM CONNECTION 5.66 (c) is the balanced equation. Mg (s) + 2 H2O (g) → Mg(OH)2 (s) + H2 (g) 5.67 (d) is the balanced equation. 2HgO → 2 Hg + O2 5.68 (c) is the balanced equation. C12H22O11 → 12 C + 11 H2O 5.69 The coefficient before O2 is (b) 2 in the balanced equation, as shown below: CH4 + 2 O2 (g) → CO2 + 2 H2O 5.70 The coefficient before ammonia, NH3, is (b) 2 in the balanced equation, as shown below: ⎯⎯
→ 2 NH3 (g) N2 (g) + 3 H2 (g) ←⎯
⎯
5.71 The oxidation number of sodium in NaI and NaNO3 is (a) +1. 108 Chapter 5 5.72 The oxidation number for nitrogen in HNO3 is (b) +5. H = +1
N=?
O = −2
HNO 3 = 0
( +1) + ?+ 3 ( −2 ) = 0
? = +5
5.73 2‐
The oxidation number for sulfur in SO4 is (c) +6. S=?
O = −2
SO 24 − = −2
?+ 4 ( −2 ) = −2
? = +6
5.74 The oxidation number for sulfur in Na2S2O3 is (c) +2. Na = +1
S=?
O = −2
Na 2 S 2 O 3 = 0
2 ( +1) + 2 ( ? ) + 3 ( −2 ) = 0
2 ( ? ) = +4
? = +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). ‐
‐
2‐
8 H+ (aq) + 6 Cl‐ (aq) + Sn (s) + 4 NO3 (aq)→ SnCl6 (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. Chemical Reactions 109 5.84 The net ionic equation is (b) Ca2+ (aq) + 2 Cl‐ (aq) → CaCl2 (s). 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 ⎯⎯
→ 2 NH3 (g) shows that (b) 3 moles of hydrogen The chemical reaction: N2 (g) + 3 H2 (g) ←⎯
⎯
gas are needed to react with one mole of nitrogen. 5.86 (c) 53.5 g NH4Cl can be produced. NH 3 + HCl → NH 4 Cl
⎛ 1 mole NH 3
17 g NH 3 ⎜⎜
⎝ 17.0 g NH 3
⎞
⎛ 1 mole HCl ⎞
36.5 g HCl ⎜⎜
⎟⎟ = 1.0 mole NH 3
⎟⎟ = 1.0 mole HCl ⎠
⎝ 36.5 g HCl ⎠
17 g NH 3 + 36.5 g HCl = 53.5 g NH 4 Cl
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. Mg + 2H + → Mg 2 + + H 2
⎛ 2 g H2 ⎞
6 g Mg ⎜⎜
⎟⎟ = 0.5 g H 2
⎝ 24 g Mg ⎠
5.88 (d) 0.25 mole of CaCl2 is needed to form 0.5 mole of NaCl. CaCl 2 + Na 2 CO 3 → CaCO 3 + 2NaCl
⎛ 1 mole CaCl 2 ⎞
0.5 mole NaCl ⎜⎜
⎟⎟ = 0.25 mole CaCl 2
⎝ 2 moles NaCl ⎠
⎛ 1 mole Na 2 CO 3
0.5 mole NaCl ⎜⎜
⎝ 2 moles NaCl
⎛ 1 mole CaCO 3
0.5 mole NaCl ⎜⎜
⎝ 2 moles NaCl
⎞
⎟⎟ = 0.25 mole Na 2 CO 3
⎠
⎞
⎟⎟ = 0.25 mole CaCO 3
⎠
5.89 In the reaction: 4Al + 3O2 → 2Al2O3, (b) 36 grams O2 are needed to completely react with 1.5 moles of Al. ⎛ 96 g O 2 ⎞
1.5 moles Al ⎜⎜
⎟⎟ = 36 g O 2 ⎝ 4 moles 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. 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 110 Chapter 5 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. 5.94 2 Cr2O7 (aq) + 3 CH3CH2OH (aq) + 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‐ 2‐
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 c. Zn + 2HCl → H2 + ZnCl2 b. 2H2O + C → CO + 2H2 d. CO + O2 → CO2 Answer: C
Chemical Reactions 111 3. Which set of coefficients balances the equation? __________ CH4 → __________C3H8 + __________ H2 a. 3,1,1 b. 3,2,1 c. 3,1,2 d. 6,2,2 Answer: C
4. When the equation below is properly balanced, what coefficient is in front of KCl? KClO3 → KCl + O2 a. 1 b. 2 c. 3 d. 4 Answer: B
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? C3H8 + O2 → CO2 + H2O a. 1 b. 2 c. 3 d. 6 Answer: C 6. Consider the reactants and propose the right side of the equation. Mg + H2SO4 → __________ a. no reaction b. MgHSO4 c. MgSO4 + H2 d. MgH2 + SO4 Answer: C
7. Consider the reactants in the partial equation given. Which choice is a product? H3PO4 + Sr(OH)2 → __________ a. no reaction b. SrH3PO5 c. SrH5PO6 d. Sr3(PO4)2 Answer: D
8. Which species would be considered a spectator ion based on the following equation? 2+
‐
+
2+
+
‐
Ba + 2NO3 + 2Na + SO4 → BaSO4(s) + 2NO3 + 2Na 2+
2+
a. Ba b. SO4 c. BaSO4 ‐
d. 2NO3 Answer: D
9. A redox reaction can also be a: a. combination reaction. b. single replacement reaction. c. decomposition reaction. d. more than one response is correct Answer: D
10. Which of the following is typically an oxidizing agent? a. oxygen b. hydrogen c. water d. hydrogen chloride Answer: A
11. Single replacement reactions are always: a. redox reactions. b. nonredox reactions. c. combination reactions. d. decomposition reactions. Answer: A
12. Which of the following equations (not balanced) represents an oxidation‐reduction process? a. AgNO3 + NaCl → AgCl + NaNO3 c. C2H4 + O2 → CO2 + H2O b. C + O2 → CO2 d. more than one response is correct Answer: D
112 Chapter 5 13. What is the oxidation number of Cl in HClO3? a. +2 b. +3 c. +4 d. +5 Answer: D
14. What is the oxidation number of Br in BrO ? a. ‐1 b. +1 c. +2 d. +3 15. What is the oxidation number of Mn in KMnO4? a. +9 b. +7 c. +5 d. +4 Answer: D
Answer: B
16. In which of the following does an element have an oxidation number of +4? a. TiO2 b. HBr c. SO3 d. SO Answer: A
17. Which of the following contains the metal with the highest oxidation number? a. CaCl2 b. NaCl c. FeCl3 d. CuCl2 Answer: Reaction: C
The following questions are based on the following reaction. 4HCl + MnO2 → Cl2 + 2H2O + MnCl2
18. Refer to Reaction. What substance is oxidized? a. Cl in HCl b. Mn in MnO2 c. H in HCl d. O in MnO2 Answer: A
19. Refer to Reaction. What substance is reduced? a. Cl in HCl b. Mn in MnO2 c. H in HCl d. O in MnO2 Answer: B
20. Refer to Reaction. What is the oxidizing agent? a. HCl b. Cl2 c. MnO2 d. MnCl2 Answer: C
21. Refer to Reaction. What is the reducing agent? a. HCl b. Cl2 c. MnO2 Answer: A
22. Which of the following is a decomposition reaction? a. SO2 + O2 → 2SO3 b. 2C2H2 + 5O2 → 4CO2 + 2H2O c. 2H2O2 → 2H2O + O2 d. AgNO3 + NaCl → AgCl + NaNO3 Answer: C
d. MnCl2 Chemical Reactions 113 23. Which of the following is a combination reaction? a. SO2 + O2 → 2SO3 c. 2H2O2 → 2H2O + O2 b. 2C2H2 + 5O2 → 4CO2 + 2H2O d. AgNO3 + NaCl → AgCl + NaNO3 Answer: A
24. Which of the following is a single replacement reaction? a. CuO + H2 → Cu + H2O c. SO2 + H2O → H2SO3 b. HBr + KOH → H2O + KBr d. 2HI → I2 + H2 Answer: A 25. Which of the following is a double replacement reaction? a. CuO + H2 → Cu + H2O c. SO2 + H2O → H2SO3 b. HBr + KOH → H2O + KBr d. 2HI → I2 + H2 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 b. a full equation c. an oxidation‐reduction equation d. a net ionic equation Answer: D
27. Which ions in the following reaction would be classified as spectator ions? Zn + 2H+ + 2Cl‐ → Zn2+ + 2Cl‐ + H2 a. Zn2+ c. Cl‐ b. H+ d. more than one response is correct. Answer: C 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 2
a. Ba + and CO 2+
+
c. Ba , CO , and H +
b. H and SO 2+
+
d. Ba , CO , H , and SO Answer: D
29. For the reaction of HCl and NaOH, which of the following type of chemicals will appear in the net equation? a. weak acid b. weak base c. salt d. water Answer: D
30. In the equation for an endothermic reaction, the word ʺheatʺ appears: a. on the right side. b. on the left side. c. on both sides. d. on neither side. Answer: B
31. The energy involved in chemical reactions: a. always appears as heat. b. always appears as light. Answer: D
c. always appears as electricity. d. can take many forms. 114 Chapter 5 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 b. 21.6 c. 145 d. 107 Answer: A
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 b. 1.5 c. 2.0 d. 3.0 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? a. 1 b. 2 c. 3 d. 6 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 b. 6.00 c. 14.0 d. 28.0 Answer: C
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? b. 3.01 x 1022 c. 6.02 x 1022 d. 2.00 a. 12.0 x 1023 Answer: C
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 c. CO d. CO and O2 Answer: C
40. Acetylene, C2H2, burns according to the following reaction: C2H2+5O2 → 4CO2+2H2O
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? a. CO2 and H2 c. C2H2, CO2, and H2O b. O2, CO2, and H2O d. O2, C2H2, CO2, and H2O Answer: C
Chemical Reactions 115 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 c. 8.06 d. 4.03 Answer: D 42. Tincture of iodine is an antiseptic. What is a tincture? a. iodine dissolved in water c. a 3% hydrogen peroxide solution b. a substance dissolved in alcohol d. an oxidizing antiseptic Answer: B
43. What coefficients are needed to balance the following reaction: CH4 + ______ O2 → ______ CO2 + ______ H2O a. 4,1,2 b. 2,0.5,1 c. 2,1,2 d. none of these Answer: C
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. c. rearranged to form new substances. b. reaction of starting materials. d. reduction in total mass. Answer: D
45. Which of the following is not a common use of the term oxidation? a. to gain electrons c. to combine with oxygen b. to lose hydrogen d. to increase in oxidation number Answer: A
46. The modern definition of oxidation is a process: a. involving a reaction with oxygen. c. in which electrons are gained. b. in which hydrogen is gained. d. none of the above Answer: D
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 % c. 94 % b. 93.86 % d. 0.9386 % Answer: B
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 c. reaction did not go to completion b. poor laboratory technique d. all of the above Answer: D
116 Chapter 5 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 c. combination b. double replacement d. decomposition Answer: A
TRUE‐FALSE 1. The reaction Mg + 3N2 → Mg3N2 is correctly balanced as written.
Answer: F
2. The reaction Ca + Cl2 → CaCl2 is a redox reaction.
Answer: 3.
T
Redox reactions may also be decomposition reactions. 4.
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 H2SO4 + 2 NaOH → Na2SO4 + 2 H2O is H+ + OH→H2O.
Answer: T
11. The oxidation number of oxygen in H2O2 is ‐2.
Answer: F
Chemical Reactions 117 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) → H
2O (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