1. Ch 4. Chemical Quantities and
Aqueous Reactions

2. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
1 mol
2 mol
Stoichiometry of the reaction
FIXED ratio for each reaction
If the amount of one chemical is known we can calculate
how much other chemicals are required or produced.

3. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
1 mol
2 mol
2 mol
4 mol
2 mol
4 mol
3 mol
6 mol
3 mol
6 mol
5.22 mol
10.44 mol
5.22 mol
10.44 mol

4. 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
6.02 mol of NH3 is used in the above reaction.
How many moles of O2 is required to react with all the NH3?
How many moles of H2O will be produced?

5. 2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g)
2 mol
3 mol
1 mol
3 mol
3 mol
6.02 mol x y z u

6. 2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g)
2 mol
3 mol
1 mol
3 mol
3 mol
6.04 g
mass?
6.04 g ÷ (14.01 g/mol x g/mol x 3) = mol
0.355 mol x y
Mass of CuO = mol x g/mol = 42.4 g
Mass of N2 = mol x g/mol = 4.99 g

7. Other Examples: page 143 ― 144

9. CH4 + 2O2 → CO2 + 2H2O 0 mol 1 mol 2 mol initial: 1 mol 2 mol final:
The actual amount of reactants consumed and actual amount
of products produced agree with the stoichiometry.
1 mol
0 mol
initial:
? mol
final:

10. CH4 + 2O2 → CO2 + 2H2O 1 mol 0 mol 1 mol x = 0.5 mol y z (1 − 0.5) mol
initial:
consumed:
1 mol x
= 0.5 mol
final: y z
(1 − 0.5) mol
0 mol
= 0.5 mol
= 1 mol
x = 0.5 mol
1 mol CH4 requires 2 mol O2, available O2 is 1 mol: limiting reagent.
Result: 1 mol O2 will be consumed completely.
CH4 will have leftover: excess reagent.

11. The reactant of which there are fewer moles than the
stoichiometry requires is the limiting reagent.
The reactant of which there are more moles than the
stoichiometry requires is the excess reagent.
Chemical reactions always occur according to the
stoichiometry, therefore the limiting reagent is consumed
and the excess reagent has leftover. The amount of products
is determined by the amounts of reagents that are actually
consumed.

12. CH4 + 2O2 → CO2 + 2H2O 1 mol 0 mol 0.5 mol 1 mol (1 − 0.5) mol 0 mol
excess reagent
limiting reagent
CH O2 → CO H2O
1 mol
0 mol
initial:
consumed:
0.5 mol
1 mol
(1 − 0.5) mol
final:
0 mol
0.5 mol
1 mol
0.5 : : : 1
1 : : : 2 =

13. + + + +

14. 2 slices of bread + 1 slice if ham → 1 sandwich
1 sandwich + 2 slices of bread
excess reagent
limiting reagent
amount of product
excess reagent
leftover

15. NH3(g) + CuO(s) → N2(g) + Cu(s) + H2O(g)
For the following reaction, if a sample containing 18.1 g of NH3
reacted with 90.4 g of CuO, which is the limiting reagent? How
many grams of N2 will be formed?
NH3(g) + CuO(s) → N2(g) + Cu(s) + H2O(g)
How many grams of excess reagent will be leftover?
If 6.63 g of N2 is actually produced, what is the percent yield?

16. Procedure for limiting/excess reagent calculations
aA + bB → cC + dD
Make sure the equation is balanced.
Find the moles of each reactant:
moles = mass in gram / molar mass
3) Pick up any reactant, say A, and use the stoichiometry to
calculate the required amount of the other reactant B.
Compare the required amount of B with the available
amount of B.
a) If required > available, then B is the limiting reagent and A
is the excess reagent.
b) If required < available, then B is the excess reagent and A
is the limiting reagent.
Use the amount of the limiting reagent and the stoichiometry
to calculate the amount of any product and the amount of the
excess reagent that has been consumed.
Leftover excess reagent = available − consumed
If actual yield is given
percent yield = (actually yield / theoretical yield) x 100%

17. Problem Set 7

19. Review experiment 11
Review problem sets 6 & 7

22. Midterm Exam Time: next week during lab session
Material covered: up to this point
Problem sets are very helpful

23. Classification of Matter
Homogeneous
(visibly indistinguishable)
Heterogeneous
(visibly distinguishable)
(Solutions)
Mixtures
(multiple components)
Pure Substances
(one component)
Matter
Elements
Compounds

24. Solute + Solvent = Solution
Solvent = water, aqueous solution
Water can dissolve many substances

25. H2O O H H
Chapter 4, Figure 4.8
Charge Distribution in a Water Molecule

26. Chapter 4, Figure 4.9
Solute and Solvent Interactions in a Sodium Chloride Solution

27. Solution conducts electricity well
Chapter 4, Figure 4.10
Sodium Chloride Dissolving in Water
Solution conducts electricity well

28. C12H22O11
Chapter 4, Figure 4.12
Sugar and Water Interactions

29. Solution does not conduct electricity
Chapter 4, Figure 4.13
A Sugar Solution
Solution does not conduct electricity

30. Solution conducts electricity, but weakly
Chapter 4, Unnumbered Figure 3, Page 148
Solution conducts electricity, but weakly

31. Based on the electrical conductivity in aqueous solution
strong electrolytes
weak electrolytes
electrolytes
nonelectrolytes
solutes

32. strong electrolytes: dissociate 100 % into ions
weak electrolytes: only a small fraction dissociate
into ions
nonelectrolytes: no dissociation

33. Chapter 4, Figure 4.14
Electrolytic Properties of Solutions

34. strong acids: HCl, HNO3, H2SO4, HClO4 strong bases: NaOH, KOH
salts: NaCl, K2SO4, ……
strong acids: HCl, HNO3, H2SO4, HClO4
strong bases: NaOH, KOH
Strong electrolytes
Bases: compounds that give OH− when dissolved in water.
weak acids: acetic acid: HC2H3O2
weak bases: ammonia: NH3
Weak electrolytes

35. Reaction of NH3 in Water

36. concentrations
no unit

37. 10. g of sugar is dissolved in 40. g of water.
What is the mass percent of sugar in this solution?

38. Unit: mol/L or M

39. Example 4.5, page 153
25.5 g of KBr is dissolved in water and forms a
solution of 1.75 L. What is the molarity of the solution?
Example 4.6, page 154
How many liters of a mol/L NaOH solution
contains mol of NaOH?

40. How to prepare 1.00 L of NaCl aqueous
solution with a molarity of 1.00 mol/L?
1.00 mol NaCl L of H2O = 1.00 mol/L NaCl (aq)

41. Chapter 4, Figure 4.5
Preparing a 1 Molar NaCl Solution

42. Solution Dilution
Concentrated solutions for storage, called stock solutions
stock solution + water  desired solution

43. moles of solute before dilution = moles of solute after dilution
M1V1 = M2V2
M1: molarity of concentrated solution
V1: volume of concentrated solution
M2: molarity of diluted solution
V2: volume of diluted solution
Example on page 155
A lab procedure calls for 3.00 L of a mol/L CaCl2
solution. How should we prepare it from a 10.0 mol/L stock
solution?

44. Chapter 4, Figure 4.6
Preparing 3.00 L of M CaCl2 from a 10.0 M Stock Solution

45. Example 4.7, page 156
To what volume should you dilute L of a 15.0 mol/L
NaOH solution to obtain a 3.00 mol/L NaOH solution?

47. Types of reactions
Precipitation reactions

48. complete ionic equation
NaCl(aq) + AgNO3(aq) 
AgCl(s) + NaNO3(aq)
formula equation
Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq)  AgCl(s) + Na+(aq) + NO3−(aq)
complete ionic equation
Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq)  AgCl(s) + Na+(aq) + NO3−(aq)
spectator ions
Cl−(aq) + Ag+(aq)  AgCl(s)
net ionic equation

50. Predict whether each compound is soluble or insoluble.
EXAMPLE 4.9 Predicting whether an Ionic Compound Is Soluble
Predict whether each compound is soluble or insoluble.
(a) PbCl2 (b) CuCl2 (c)Ca(NO3)2 (d) BaSO4

52. BaCl2(aq)  Ba2+(aq) + 2Cl−(aq)
BaCl2(aq) + K2SO4(aq) 
BaSO4(s) + 2KCl(aq)
BaCl2(aq)  Ba2+(aq) + 2Cl−(aq)
K2SO4(aq)  2K+ (aq) + SO42− (aq)
Ba2+(aq) + 2Cl−(aq) + 2K+ (aq) + SO42− (aq)  BaSO4(s) + 2Cl−(aq) + 2K+(aq)
Ba2+(aq) + SO42− (aq)  BaSO4(s)

53. BaCl2(aq) + KNO3(aq) 
BaCl2(aq) + 2KNO3(aq)  Ba(NO3)2(aq) + 2KCl(aq)
BaCl2(aq)  Ba2+(aq) + 2Cl−(aq)
KNO3(aq)  K+ (aq) + NO3− (aq)
2KNO3(aq)  2K+ (aq) + 2NO3− (aq)
Ba2+(aq) + 2Cl−(aq) + 2K+ (aq) + 2NO3− (aq) 
Ba2+(aq) + 2NO3− (aq) + 2Cl−(aq) + 2K+(aq)

54. Types of reactions
Precipitation reactions
Acid-base reactions

55. Acid: Substance that produces H+ ions in aqueous solution
Base: Substance that produces OH− ions in aqueous solution

56. Chapter 4, Unnumbered Figure 1, Page 156

57. Chapter 4, Unnumbered Figure 2, Page 156

58. Try to remember them

59. HCl(aq)  H+(aq) + Cl−(aq)
NaOH(aq)  Na+(aq) + OH−(aq)
H+(aq) + OH−(aq)  H2O(l)
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
H+(aq) + Cl−(aq) +Na+(aq) +OH−(aq)  H2O(l) + Na+(aq) + Cl−(aq)
H+(aq) + OH−(aq)  H2O(l)
neutralization
acidic
basic
neutral

60. What is the molarity of HCl(aq) or H+(aq)? HCl(aq)  H+(aq) + Cl−(aq)
Chapter 4, Unnumbered Figure, Page 158

61. When reaction completes
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
When reaction completes
nNaOH = nHCl
MNaOHVNaOH = MHClVHCl
prepared,
known
measured
by buret, known
unknown
measured by
pippet, known

62. MNaOHVNaOH = MHClVHCl
Chapter 4, Figure 4.19
Acid–Base Titration

63. Read acid-base titration starting on page 171.
Chapter 4, Figure 4.20
Titration
End point: light pink
Read acid-base titration starting on page 171.
Read the online instruction next week for the titration experiment.

65. MNaOHVNaOH = MHClVHCl HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
The titration of a mL sample of an HCl solution of unknown concentration requires mL of a mol/L NaOH solution to reach the equivalence point. What is the concentration of the unknown HCl solution in mol/L?
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
MNaOHVNaOH = MHClVHCl

66. Types of reactions Precipitation reactions Acid-base reactions
Oxidation-Reduction reactions

67. 2Mg(s) + O2(g) → 2MgO(s)
Reactions that involve electron transfer are called
oxidation-reduction reactions, or redox reactions.

68. Oxidation number (state)
A way to keep track of the electrons gained or lost
1) For atoms in its elemental form, oxidation number = 0
Na, Ar, O2, N2, O3, P4, S8
2) For monatomic ion, oxidation number = charge of the ion
Na+, Ca2+, Co2+, Co3+, Cl−, O2−
NaCl, Na2O, CaCl2, CaO, CoCl2, CoCl3, Co2O3, CoO

69. H2O O H H
Chapter 4, Figure 4.8
Charge Distribution in a Water Molecule

70. 3) In covalent compounds
Remember O: −2 H: +1 F: −1
In a neutral compound, the sum of the oxidation number = 0
CO, CO2, SF6, SF4, H2S, NH3, P2O5, N2O3
In a polyatomic ion, the sum of the oxidation number = ion charge
NO3−, SO42−, NH4+, Cr2O72−, MnO4−

72. Reactions that cause change of oxidation numbers
are called redox reactions.
Element loses electrons → its oxidation number increases
→ element is oxidized → oxidation reaction
Substance that contains the oxidized element is call the
reducing agent.
Element gains electrons → its oxidation number decreases
→ element is reduced → reduction reaction
Substance that contains the reduced element is call the
oxidizing agent.

73. PbO (s) + CO (g) → Pb (s) + CO2 (g)

74. Cu − 2e−  Cu2+

75. Problem Set 8 concentrations: mass percent and molarity;
precipitation reactions; redox reactions.