1. Chemical Equations Chapter 10
Eugene Passer Chemistry Department Bronx Community College
© John Wiley and Sons, Inc
Version 1.1

2. Chapter Outline 8.1 The Chemical Equation
8.4 Types of Chemical Equations
8.2 Writing and Balancing Equations
8.5 Heat in Chemical Reactions
8.3 What Information Does an Equation Tell Us

3. Chemical equations provide us with the means to
Chemists use chemical equations to describe reactions they observe in the laboratory or in nature.
Chemical equations provide us with the means to
summarize the reaction
display the substances that are reacting
show the products
indicate the amounts of all component substances in a reaction.

4. The Chemical Equation

5. Chemical reactions always involve change.
Atoms, molecules or ions rearrange to form new substances.
The substances entering the reaction are called reactants.
The substances formed in the reaction are called products.
During reactions, chemical bonds are broken and new bonds are formed.

6. A chemical equation is a shorthand expression for a chemical change or reaction.
A chemical equation uses the chemical symbols and formulas of the reactants and products and other symbolic terms to represent a chemical reaction.

7. Coefficients (whole numbers) are placed in front of substances to balance the equation and to indicate the number of units (atoms, molecules, moles, or ions) of each substance that are reacting.

8. coefficient 2
Al + Fe2O3  Fe + Al2O3

9. Conditions required to carry out the reaction may be placed above or below the arrow.

10. Al + Fe2O3  Fe + Al2O3
coefficient 2 
 heat

11. The physical state of a substance is indicated by symbols such as (l) for liquid.

12. 2Al(s) + Fe2O3(s)  2Fe(l) + Al2O3 (s) (s) (l)
In a chemical reaction atoms are neither created nor destroyed.
2Al(s) + Fe2O3(s)  2Fe(l) + Al2O3 (s)
(s)
(l)
All atoms present in the reactant must also be present in the products.

13. Symbols Used in Chemical Reactions

14. Al + Fe2O3  Fe + Al2O3 Al + Fe2O3  Fe + Al2O3 Chemical Equation
reactants
products
Al + Fe2O3  Fe + Al2O3
Al + Fe2O3  Fe + Al2O3
iron oxygen bonds break
aluminum oxygen bonds form

15. symbol
(aq)
aqueous
meaning
after formula
location

16. placed between substances
symbol +
meaning
plus
placed between substances
location

17. symbol 
heat
meaning
written above 
location

18. symbol 
gas formation
meaning
after formula
location

19. Writing and Balancing Equations

20. Never change a correct formula to balance an equation.
To balance an equation adjust the number of atoms of each element so that they are the same on each side of the equation.
Never change a correct formula to balance an equation.

21. Steps for Balancing Equations

22. mercury(II) oxide → mercury + oxygen
Step 1 Identify the reaction. Write a description or word equation for the reaction.
Mercury (II) oxide decomposes to form mercury and oxygen.
mercury(II) oxide → mercury + oxygen

23. The formulas of the reactants and products can never be changed.
Step 2 Write the unbalanced (skeleton) equation.
The formulas of the reactants and products must be correct.
The reactants are written to the left of the arrow and the products to the right of the arrow.
HgO  Hg + O2
The formulas of the reactants and products can never be changed.

24. Element Reactant Side Product Side
Step 3a Balance the equation.
HgO → Hg + O2
Element Reactant Side Product Side Hg
There is one mercury atom on the reactant side and one mercury atom on the product side.
Mercury is balanced.

25. Step 3a Balance the equation.
Count and compare the number of atoms of each element on both sides of the equation.
Determine the elements that require balancing.

26. Element Reactant Side Product Side
Step 3a Balance the equation.
HgO  Hg + O2
Element Reactant Side Product Side O
There are two oxygen atoms on the product side and there is one oxygen atom on the reactant side.
Oxygen needs to be balanced.

27. Step 3b Balance the equation.
Balance each element one at a time, by placing whole numbers (coefficients) in front of the formulas containing the unbalanced element.
A coefficient placed before a formula multiplies every atom in the formula by that coefficient.

28. Element Reactant Side Product Side
Step 3b Balance the equation.
HgO  Hg + O2 2
Element Reactant Side Product Side O
Place a 2 in front of HgO to balance O.
There are two oxygen atoms on the reactant side and there are two oxygen atoms on the product side.
Oxygen (O) is balanced.

29. Step 3c Balance the equation.
Check all other elements after each individual element is balanced to see whether, in balancing one element, another element became unbalanced.

30. Element Reactant Side Product Side
Step 3c Balance the equation.
2HgO  Hg + O2
Element Reactant Side Product Side Hg
Count and compare the number of mercury (Hg) atoms on both sides of the equation.
There are two mercury atoms on the reactant side and there is one mercury atom on the product side.
Mercury (Hg) is not balanced.

31. Element Reactant Side Product Side
Step 3c Balance the equation.
2HgO  Hg + O2 2
Element Reactant Side Product Side Hg
Place a 2 in front of Hg to balance mercury.
There are two mercury atoms on the reactant side and there are two mercury atoms on the product side.
Mercury (Hg) is balanced.

32. Element Reactant Side Product Side
 THE EQUATION IS BALANCED
2HgO  2Hg + O2
Element Reactant Side Product Side Hg O

33. sulfuric acid + sodium hydroxide → sodium sulfate + water
Balance the Equation
sulfuric acid + sodium hydroxide → sodium sulfate + water

34. H2SO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l)
Balance the Equation
H2SO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l) 2
Reactant Side Product Side
SO4 1 1
Na 1 2
O 1 1
H 3 2 2 2 4
Place a 2 in front of NaOH to balance Na.
There is one Na on the reactant side and there are two Na on the product side.

35.  THE EQUATION IS BALANCED 
H2SO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l) 2 2
Reactant Side Product Side
SO4 1 1
Na 2 2
O 2 1
H 4 2 2 4
Place a 2 in front of H2O to balance H.
There are 4 H on the reactant side and two H on the product side.

36. butane + oxygen → carbon dioxide + water
Balance the Equation
butane + oxygen → carbon dioxide + water

37.  THE EQUATION IS BALANCED 
C4H10 (g) + O2 (g) → CO2(g) + H2O(l) 2 8 10 13
Reactant Side Product Side
C 8 8
H 20 20 O 26
Place a 13 in front of O2 to balance O.
There are now 26 O on the product side.

38. What Information Does an Equation Tell Us

39. The meaning of a formula is context dependent.
The formula H2O can mean:
2H and 1 O atom
1 molecule of water
1 mol of water
6.022 x 1023 molecules of water
18.02 g of water

40. In an equation formulas can represent units of individual chemical entities or moles.+
Cl2
2HCl →
1 molecule H2
1 molecule Cl2
2 molecules HCl
1 mol H2
1 mol Cl2
2 mol HCl

41. Formulas
Number of molecules
Number of atoms
Number of moles
Molar masses

42. Types of Chemical Equations

43. Combination
Decomposition
Single-Displacement
Double-Displacement
Combustion

44. Combination Reactions

45. Two reactants combine to form one product.
A + B  AB

46. Metal + Oxygen → Metal Oxide
2Ca(s) + O2(g)  2CaO(s)
4Al(s) + 3O2(g)  2Al2O3(s)

47. Nonmetal + Oxygen → Nonmetal Oxide
S(s) + O2(g)  SO2(g)
N2(g) + O2(g)  2NO(g)

48. 2K(s) + F2(g)  2KF(s) 2Al(s) + 3Cl2(g)  2AlCl3(s)
Metal + Nonmetal → Salt
2K(s) + F2(g)  2KF(s)
2Al(s) + 3Cl2(g)  2AlCl3(s)

49. Na2O(s) + H2O(l)  2NaOH(aq)
Metal Oxide + Water → Metal Hydroxide
Na2O(s) + H2O(l)  2NaOH(aq)
CaO(s) + 2H2O(l)  2Ca(OH)2(aq)

50. Nonmetal Oxide + H2O(l) → Oxy-acid
SO3(g) + H2O(l)  H2SO4(aq)
N2O5(g) + H2O(l)  2HNO3(aq)

51. Decomposition Reactions

52. A single substance breaks down to give two or more different substances.
AB  A + B

53. CaCO3(s)  CaO(s) + CO2(g)
Carbonate → CO2(g)
CaCO3(s)  CaO(s) + CO2(g)
Hydrogen Carbonate → CO2(g)
2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)

54. 2Ag2O(s)  4Ag(s) + O2(g) 2PbO2(s)  2PbO(s) + O2(g)
Metal Oxide → Metal + Oxygen
2Ag2O(s)  4Ag(s) + O2(g)
Metal Oxide → Metal Oxide + Oxygen
2PbO2(s)  2PbO(s) + O2(g)

55. 2NaNO3(s)  2NaNO2(s) + O2(g)
Miscellaneous Reactions
2H2O2(l)  2H2O(l) + O2(g)
2KClO3(s)  2KCl(s) + 3O2(g)
2NaNO3(s)  2NaNO2(s) + O2(g)

56. Single Displacement Reactions

57. One element reacts with a compound to replace one the elements of that compound.
A + BC  AC + B

58. Metal + Acid → Hydrogen + Salt
Mg(s) + 2HCl(aq)  H2(g) + MgCl2(aq)
salt
2Al(s) + 3H2SO4(aq)  3H2(g) + Al2(SO4)3(aq)
salt

59. 2Na(s) + 2H2O(l)  H2(g) + 2NaOH(aq)
Metal + Water → Hydrogen + Metal Hydroxide
2Na(s) + 2H2O(l)  H2(g) + 2NaOH(aq)
metal hydroxide
Ca(s) + 2H2O(l)  H2(g) + Ca(OH)2(aq)
metal hydroxide

60. 3Fe(s) + 4H2O(g)  4H2(g) + Fe3O4(s)
Metal + Water → Hydrogen + Metal Oxide
3Fe(s) + 4H2O(g)  4H2(g) + Fe3O4(s)
metal oxide

61. The Activity Series

62. Metals K Ca Na Mg Al Zn Fe Ni Sn Pb H Cu Ag Hg
An atom of an element in the activity series will displace an atom of an element below it from one of its compounds .
Metals K Ca Na Mg Al Zn Fe Ni Sn Pb H Cu Ag Hg
Sodium (Na) will displace an atom below it from one of its compounds.
increasing activity

63. Mg(s) + PbS(s)  MgS(s) + Pb(s)
Metal Higher in Activity Series Displacing Metal Below It
Mg(s) + PbS(s)  MgS(s) + Pb(s)
Metals Mg Al Zn Fe Ni Sn Pb
Magnesium is above lead in the activity series.

64. Ag(s) + CuCl2(s)  no reaction
Metal Lower in Activity Cannot Displace Metal Above It
Ag(s) + CuCl2(s)  no reaction
Metals Pb H Cu Ag Hg
Silver is below copper in the activity series.

65. Double Displacement Reactions

66. Cl2(g) + CaBr2(s)  CaCl2(aq) + Br2(aq)
Halogen Higher in Activity Series Displaces Halogen Below It
Cl2(g) + CaBr2(s)  CaCl2(aq) + Br2(aq)
Halogens F2 Cl2 Br2 I2
Chlorine is above bromine in the activity series.

67. AB + CD  AD + CB A displaces C and combines with D
The reaction can be thought of as an exchange of positive and negative groups.
Two compounds exchange partners with each other to produce two different compounds.
B displaces D and combines with C
AB + CD  AD + CB

68. The Following Accompany Double Displacement Reactions
formation of a precipitate
release of gas bubbles
release of heat
formation of water

69. Acid Base Neutralization
acid + base → salt + water
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(l)

70. Formation of an Insoluble Precipitate
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq)

71. CaO(s) + 2HCl(aq)  CaCl2(s) + H2O(l)
Metal Oxide + Acid
metal oxide + acid → salt + water
CuO(s) + 2HNO3(aq)  Cu(NO3)2(aq) + H2O(l)
CaO(s) + 2HCl(aq)  CaCl2(s) + H2O(l)

72. indirect gas formation
Formation of a Gas
H2SO4(aq) + 2NaCN(aq)  Na2SO4(aq) + 2HCN(g)
NH4Cl(aq) + NaOH(aq)  NaCl(aq) + NH4OH(aq)
indirect gas formation
NH4OH(aq)  NH3(g) + H2O(l)

73. Introduction to Stoichiometry: The Mole-Ratio Method

74. Stoichiometry: The area of chemistry that deals with the quantitative relationships between reactants and products.
Mole Ratio: a ratio between the moles of any two substances involved in a chemical reaction.
The coefficients used in mole ratio expressions are derived from the coefficients used in the balanced equation.

75. N2 + 3H2  2NH3
1 mol
2 mol
3 mol

76. N2 + 3H2  2NH3
1 mol
2 mol
3 mol

77. The mole ratio is used to convert the number of moles of one substance to the corresponding number of moles of another substance in a stoichiometry problem.
The mole ratio is used in the solution of every type of stoichiometry problem.

78. Step 1 Determine the number of moles of starting substance.
Identify the starting substance from the data given in the problem statement. Convert the quantity of the starting substance to moles, if it is not already in moles.

79. Step 2 Determine the mole ratio of the desired substance to the starting substance.
The number of moles of each substance in the balanced equation is indicated by the coefficient in front of each substance. Use these coefficients to set up the mole ratio.

80. Step 2 Determine the mole ratio of the desired substance to the starting substance.
Multiply the number of moles of starting substance (from Step 1) by the mole ratio to obtain the number of moles of desired substance.

81. Step 3. Calculate the desired substance in the units specified in the problem.
If the answer is to be in moles, the calculation is complete
If units other than moles are wanted, multiply the moles of the desired substance (from Step 2) by the appropriate factor to convert moles to the units required.

82. Step 3. Calculate the desired substance in the units specified in the problem.

83. Step 3. Calculate the desired substance in the units specified in the problem.

84. Step 3. Calculate the desired substance in the units specified in the problem.

85. Limiting-Reactant and Yield Calculations

86. The limiting reactant limits the amount of product that can be formed.
The limiting reactant is one of the reactants in a chemical reaction.
It is called the limiting reactant because the amount of it present is insufficient to react with the amounts of other reactants that are present.
The limiting reactant limits the amount of product that can be formed.

87. Steps Used to Determine the Limiting Reactant

88. Calculate the amount of product (moles or grams, as needed) formed from each reactant.
Determine which reactant is limiting. (The reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess.
Calculate the amount of the other reactant required to react with the limiting reactant, then subtract this amount from the starting quantity of the reactant. This gives the amount of the that substance that remains unreacted.

89. Reaction Yield

90. The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation.

91. Many reactions fail to give a 100% yield of product.
This occurs because of side reactions and the fact that many reactions are reversible.

92. The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant.
The actual yield is the amount of product finally obtained from a given amount of reactant.

93. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100.