1. Theoretical Yield: Which Reactant is Limiting?
1) calculate moles (or mass) of product formed by complete reaction of each reactant.
2) the reactant that yields the least product is the limiting reactant (or limiting reagent).
3) the theoretical yield for a reaction is the maximum amount of product that could be generated by complete consumption of the limiting reagent.

2. An Ice Cream Sundae Analogy for Limiting Reactions
Fig. 3.10

3. Limiting Reactant Example 2 NH3 is the limiting reagent.
4NH3 + 5O2 → 4NO + 6H2O
Add: 14 mol mol
Could make
16 mol NO
Could make
14 mol NO
NH3 is the limiting reagent.
(Use this as basis for all further calculations)

5. Limiting Reactant Example 3
When 66.6 g of O2 gas is mixed with 27.8 g of NH3 gas and 25.1 g of CH4 gas, 36.4 g of HCN gas is produced by the following reaction:
2CH4 + 2NH3 + 3O2 → 2HCN + 6H2O
g/mol
What is the % yield of HCN in this reaction?
How many grams of NH3 remain?

6. Which reactant is limiting?
Mass to moles
66.6 g of O2 → 2.08 mol O2
27.8 g of NH3 → 1.63 mol NH3
25.1 g of CH4 → 1.56 mol CH4
Which reactant is limiting?
2.08 mol O2 can yield 1.39 mol (or 37.5 g) HCN
1.63 mol NH3 can yield 1.63 mol (or 44.1 g) HCN
1.56 mol CH4 can yield 1.56 mol (or 42.2 g) HCN
Conclusion?
O2 is the limiting reagent.

7. O2 is the limiting reagent
O2 is the limiting reagent. So, the theoretical yield is based on 100% consumption of O2.
2.08 mol O2 can yield 1.39 mol (or 37.5 g) HCN
% yield = actual yield
x 100
theoretical yield
% yield = g HCN
x 100 = 97.1%
37.5 g HCN

8. 36.4 g (or 1.35 mol) of HCN gas is produced
2. How many grams of NH3 remain?
36.4 g (or 1.35 mol) of HCN gas is produced
2CH4 + 2NH3 + 3O2 → 2HCN + 6H2O
Since the reaction stoichiometry is 1:1, 1.35 mol of NH3 is consumed:
1.63 mol NH3 initially present
– 1.35 mol NH3 consumed
0.28 mol NH3 remaining
0.28 mol NH3 x (17.03 g NH3/mol) = 4.8 g NH3 remain

9. Urea, CH4N2O, is used as a nitrogen fertilizer
Urea, CH4N2O, is used as a nitrogen fertilizer. It is manufactured from ammonia and carbon dioxide at high pressure and high temperature:
2NH3 + CO2(g)  CH4N2O + H2O
In a laboratory experiment, 10.0 g NH3 and 10.0 g CO2 were added to a reaction vessel. What is the maximum quantity (in grams) of urea that can be obtained? How many grams of the excess reactant are left at the end of the reactions? 9

10. CO2 is the limiting reactant.
Molar masses
NH3 1(14.01) + 3(1.008) = g
CO2 1(12.01) + 2(16.00) = g
CH4N2O 1(12.01) + 4(1.008) + 2(14.01) + 1(16.00) = g
CO2 is the limiting reactant.
13.6 g CH4N2O will be produced. 10

11. To find the excess NH3, we find how much NH3 reacted:
Now subtract the amount reacted from the starting amount:
10.0 at start
reacted
2.27 g remains
2.3 g NH3 is left unreacted.
(1 decimal place) 11

12. (2 significant figures)
2NH3 + CO2(g)  CH4N2O + H2O
When 10.0 g NH3 and 10.0 g CO2 are added to a reaction vessel, the limiting reactant is CO2. The theoretical yield is 13.6 of urea. When this reaction was carried out, 9.3 g of urea was obtained. What is the percent yield?
Theoretical yield = 13.6 g
Actual yield = 9.3 g
= 68% yield
(2 significant figures) 12

13. Acid-Base

15. Two Definitions of Acids & Bases
1) Arrhenius: When dissolved in water,
acids produce H HCl (aq) → H+ (aq) + Cl- (aq)
bases produce OH- NaOH (aq) → Na+ (aq) + OH- (aq)
2) Brønsted-Lowry: Proton transfer
acids are proton donors
bases are proton acceptors
NaOH (aq) + HCl (aq) → ?
Na+ (aq) + OH- (aq) + H+ (aq) + Cl- (aq) → ?
NaOH (aq) + HCl (aq) → H2O (l) + NaCl (aq)

16. Acids - A Group of Covalent Molecules Which Lose Hydrogen Ions to Water Molecules in Solution
When gaseous hydrogen iodide dissolves in water, the attraction of the
oxygen atom of the water molecule for the hydrogen atom in HI is
greater that the attraction of the of the iodide ion for the hydrogen atom,
and it is lost to the water molecule to form an hydronium ion and an
iodide ion in solution. We can write the hydrogen atom in solution as
either H+(aq) or as H3O+(aq) they mean the same thing in solution. The
presence of a hydrogen atom that is easily lost in solution is an “Acid”
and is called an “acidic” solution. The water (H2O) could also be written
above the arrow indicating that the solvent was water in which the HI
was dissolved.
HI(g) + H2O(L) H+(aq) + I -(aq)
HI(g) + H2O(L) H3O+(aq) + I -(aq)
H2O
HI(g) H+(aq) + I -(aq)

18. Figure 4. 8B: Red cabbage juice added to solutions in the beakers
Figure 4.8B: Red cabbage juice added to solutions in the beakers. Photo courtesy of James Scherer.

19. Molecular representation of ammonium hydroxide.
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)

20. Reaction of nitric acid with water.
HNO3(aq)+ H2O(l)  NO3-(aq) + H3O+(aq)

22. Two Types of Acid-Base Reactions
“salt”
water
1) A-B Neutralization:
LiOH (aq) + HCN (aq) → LiCN (aq) + H2O (l)
H2SO4 (aq) + Ca(OH)2 (aq) → ?
gas
2) A-B Reactions with Gas Formation:
Na2CO3 (aq) + 2HBr (aq) → 2NaBr (aq) + H2O (l) + CO2 (g)
Li2SO3 (aq) + NaOH (aq) → ?
Which salts? carbonates  CO2
sulfites  SO2
sulfides  H2S

23. Beaker with Na+(aq), C2H302-(aq), and SrS04 (solid).
Na2SO4 (aq)+ Sr(C2H3O2)2(aq)  SrSO4 (s)+ NaC2H3O2 (aq)
2 Na+(aq) + SO4-2 (aq)+ Sr+2(aq) + 2C2H3O2 -(aq)
 SrSO4 (s)+ 2 Na+(aq) + 2C2H3O2 -(aq)
SO4-2 (aq)+ Sr+2(aq) SrSO4 (s)

24. Figure 4. 9: Reaction of a carbonate with an acid
Figure 4.9: Reaction of a carbonate with an acid. Photo courtesy of American Color.

27. Oxidation-Reduction
2Na (s) + Cl2(g) 2NaCl(s)

28. Figure 4. 10: Iron nail and copper ( II) sulfate
Figure 4.10: Iron nail and copper ( II) sulfate. Photo courtesy of American Color.

29. Figure 4.10: Fe reacts with Cu2+(aq) and makes Cu(s).

30. Figure 4.10: The copper metal plates out on the nail.
Write a net ionic equation for this reaction!
Cu+2(aq) + Fe(s)  Cu(s) + Fe+2(aq)

31. Redox: changing oxidation numbers
a real charge (ionic compounds) or hypothetical charge (molecular compounds, polyatomic ions) associated with an individual atom in a compound.
The oxidation number allows for electron accounting
N2 (g) + 3H2 (g) → 2NH3 (g)
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

32. Rules for Assigning Oxidation Numbers

33. = M Molar concentration “Molarity” or “M” mol of solute
liter of solution
NOT:
mol solute
liter of solvent
= M L
mark

34. Tips for Molarity-Based Calculations
Use molarity to convert volume of solution to moles of solute: Mi × Vi = mol solute
Use Mi × Vi = Mf × Vf to calculate concentrations of solutions after dilution
Never use this for reactions (e.g. neutralization)
Use Mi × Vi = mol solute & stoichiometry to calculate concentrations of sample solutions in reactions (e.g. titrations).

35. Molarity (Concentration of Solutions)= M
Moles of Solute Moles
Liters of Solution L
M = =
solute = material dissolved into the solvent
In air , Nitrogen is the solvent and oxygen, carbon dioxide, etc.
are the solutes.
In sea water , Water is the solvent, and salt, magnesium chloride, etc.
In brass , Copper is the solvent (90%), and Zinc is the solute(10%)

36. Fig. 3.11

37. Preparing a Solution - I
Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to ml or l !
What is the Molarity of the salt and each of the ions?
Na3PO4 (s) + H2O(solvent) = 3 Na+(aq) + PO4-3(aq)

38. Preparing a Solution - II
Mol wt of Na3PO4 = g / mol
3.95 g / g/mol = mol Na3PO4
dissolve and dilute to ml
M = mol Na3PO4 / l = M Na3PO4
for PO4-3 ions = M
for Na+ ions = 3 x M = M

39. Stoichiometry & Ion Dissociation
Concept Check
Stoichiometry & Ion Dissociation
For a 0.27 M aq. solution of sodium carbonate:
Write the dissociation reaction & identify solute(s).
Find the molarity of Na+ (aq) & CO32- (aq).
If you had 185 mL of this solution, how many moles of Na+ (aq) & CO32- (aq) would be present?
If you added excess MgBr2 (aq), would you expect a rxn.? If so, how many moles of solid would form?

40. Converting a Concentrated Solution to a Dilute Solution

41. Mass of solute does not change!
i = initial
f = final
Add more solvent
(“Dilution”)
Mass of solute does not change!
mol solute
liter of solution x
liter of sample =
Mi x Vi = mol solute = Mf x Vf
Constant!

42. The Dilution Dogma:
NEVER FORGET IT!
M1V1=M2V2

43. Dilution of Solutions Take 25.00 ml of the 0.0400 M KMnO4
Dilute the ml to l - What is the resulting Molarity of the diluted solution?
# moles = Vol x M
l x M = Moles
Mol / 1.00 l = M

44. How much 0.20 M HCl is needed to make
Concept Check
Dilution Calculation
How much 0.20 M HCl is needed to make
50 mL of 10 mM HCl solution?
Mi × Vi = Mf × Vf
(0.20 M HCl) × ( L) = (0.010 M HCl) × (0.050 L)
 (L) = (0.010 M HCl) × (0.050 L) = 2.5 × 10-3 L
0.20 M HCl

45. Dilution & Solution Examples
Concept Check
Dilution & Solution Examples
A) We have a 3.0 M aqueous solution of H2SO4.
How do you make 100. mL of 1.4 M H2SO4(aq)?
B) Determine how you would make 250. mL of
0.56 M CaCl2(aq)? What is [Ca2+]? [Cl-]?
(Reagent is solid is CaCl2·2H20; g/mol)
C) Predict what would happen in you mixed
solutions A and B together.
H2SO4(aq) + CaCl2(aq) → ?

46. How could you make 5.0 L of 0.025 M sucrose from a
solution which is M sucrose?
Mix L of M sucrose with 3.75 L water.

47. Two Types of Acid-Base Reactions
“salt”
water
1) A-B Neutralization:
LiOH (aq) + HCN (aq) → LiCN (aq) + H2O (l)
H2SO4 (aq) + Ca(OH)2 (aq) → ?
gas
2) A-B Reactions with Gas Formation:
Na2CO3 (aq) + 2HBr (aq) → 2NaBr (aq) + H2O (l) + CO2 (g)
Li2SO3 (aq) + NaOH (aq) → ?
Which salts? carbonates  CO2
sulfites  SO2
sulfides  H2S

48. Neutralization Calculation
Concept Check
Neutralization Calculation
How much 2.0 M HCl is needed to “neutralize” 2.3 liters of 0.15 M NaOH?
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
acid base → “salt” water
or…..
How much 2.0 M HCl should be added such that the mol of H+ (aq) = mol OH– (aq)?
N.I.E.: H+ (aq) + OH- (aq) → H2O (l)

49. Acids: 1) Nitric acid 2) Carbonic acid 3) Phosphoric acid
Neutralization Calculations
Determine the volume of 0.10 M KOH(aq) solution required to neutralize mL samples of three different acids, all at 0.20 M.
Acids: 1) Nitric acid 2) Carbonic acid 3) Phosphoric acid
1) __ HNO3 (aq) + __ KOH (aq) →
2) __ KOH (aq) →
3) __ KOH (aq) →

50. Titration Calculation
4.49 mL of M NaOH is required to titrate a mL sample of H2SO4 to the endpoint. What is the molar concentration of H2SO4 in the sample?
NaOH (aq) + H2SO4 (aq) Na2SO4 (aq) + H2O (l)
2NaOH (aq) + 1H2SO4 (aq) Na2SO4 (aq) + 2H2O (l)
Stoichiometry not 1:1 !!!

51. Too far!
M NaOH
∆V = 4.49 mL
25.00 mL
[H2SO4] = ?
Starting
point
At equivalence
point

52. 2NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + 2H2O (l)x
L of NaOH (aq)
required to get
to endpoint
Measured
with buret
mol NaOH
liter of solution
Known =
mol NaOH
consumed in
neutralization
Calculated
mole
ratio
mol H2SO4
25.00 mL
sample
[H2SO4] (M)

53. (from chemical equation)
2NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + 2H2O (l) x
L of NaOH (aq)
required to get
to endpoint =
mol NaOH
consumed in
neutralization
liter of solution
mol/L x L = mol
1 mol H2SO4
2 mol NaOH
(from chemical equation) x
mol
NaOH consumed by neutralization
mol
H2SO4 present in
the initial sample =

54. mol H2SO4 present in initial sample
(calculated)
mol H2SO4 present in initial sample
volume initial sample =
[H2SO4] (M)
in the
initial sample =
mol
liter
L (known)
[H2SO4 (aq)] = M

56. ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)
Zinc sulfide reacts with hydrochloric acid to produce hydrogen sulfide gas:
ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)
How many milliliters of M HCl are required to react with g ZnS? 56

57. Molar mass of ZnS = g
= L = 157 mL HCl solution 57

59. Chemical Equation Calculation - III
Mass
Atoms (Molecules)
Molecular
Weight
Avogadro’s
Number
g/mol
6.02 x 1023
Molecules
Reactants
Products
Moles
Molarity
moles / liter
Solutions

60. Calculating Mass of Solute from a Given Volume of Solution
Volume (L) of Solution
Molarity M = (mol solute / Liters of solution) = M/L
Moles of Solute
Molar Mass (M) = ( mass / mole) = g/mol
Mass (g) of Solute

61. CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
2 g 10 mL
0.75 M
Which is limiting?
2 g CaCO3 x 1 mol CaCl2 = mol CaCl2
100 g CaCO mol CaCO3
0.01 L HCl x 0.75 mol HCl x 1 mol CaCl2 =
L HCl 2 mol HCl
0.004 mol CaCl2
What is the [Cl-] after
the reaction?
How many g of CaCO3 remain?

62. Calculating Amounts of Reactants and
Products for a Reaction in Solution
Al(OH)3 (s) + 3 HCl (aq) H2O(l) + AlCl3 (aq)
Given 10.0 g Al(OH)3, what volume of
1.50 M HCl is required to neutralize the
base?
Mass (g) of Al(OH)3
M (g/mol)
10.0 g Al(OH)3
78.00 g/mol Al(OH)3
Moles of Al(OH)3
molar ratio
Moles of HCl
M ( mol/L)
Volume (L) of HCl

63. Calculating Amounts of Reactants and
Products for a Reaction in Solution
Al(OH)3 (s) + 3 HCl (aq) H2O(l) + AlCl3 (aq)
Given 10.0 g Al(OH)3, what volume of
1.50 M HCl is required to neutralize the
base?
Mass (g) of Al(OH)3
M (g/mol)
10.0 g Al(OH)3
78.00 g/mol
= mol Al(OH)3
Moles of Al(OH)3
0.128 mol Al(OH)3 x =
molar ratio
Moles HCl
Moles of HCl
M ( mol/L)
Volume (L) of HCl

64. Calculating Amounts of Reactants and
Products for a Reaction in Solution
Al(OH)3 (s) + 3 HCl (aq) H2O(l) + AlCl3 (aq)
Given 10.0 g Al(OH)3, what volume of
1.50 M HCl is required to neutralize the
base?
Mass (g) of Al(OH)3
M (g/mol)
10.0 g Al(OH)3
78.00 g/mol
= mol Al(OH)3
Moles of Al(OH)3
3 moles HCl
moles Al(OH)3
0.128 mol Al(OH)3 x =
molar ratio
0.385 Moles HCl
Moles of HCl
Vol HCl
M ( mol/L)
Volume (L) of HCl

65. Calculating Amounts of Reactants and
Products for a Reaction in Solution
Al(OH)3 (s) + 3 HCl (aq) H2O(l) + AlCl3 (aq)
Given 10.0 g Al(OH)3, what volume of
1.50 M HCl is required to neutralize the
base?
Mass (g) of Al(OH)3
M (g/mol)
10.0 g Al(OH)3
78.00 g/mol
= mol Al(OH)3
Moles of Al(OH)3
3 moles HCl
moles Al(OH)3
0.128 mol Al(OH)3 x =
molar ratio
0.385 Moles HCl
Moles of HCl
1.00 L HCl
1.50 Moles HCl
Vol HCl = x Moles HCl
Vol HCl = L = 256 ml
M ( mol/L)
Volume (L) of HCl

66. Solving Limiting Reactant Problems in
Solution - Precipitation Problem - I
Problem: Lead has been used as a glaze for pottery for years, and can be
a problem if not fired properly in an oven, and is leachable from the
pottery. Vinegar is used in leaching tests, followed by Lead precipitated
as a sulfide. If ml of a M solution of Lead nitrate is added
to ml of a M solution of Sodium sulfide, what mass of
solid Lead Sulfide will be formed?
Plan: It is a limiting-reactant problem because the amounts of two
reactants are given. After writing the balanced equation, determine the
limiting reactant, then calculate the moles of product. Convert moles of
product to mass of the product using the molar mass.
Solution: Writing the balanced equation:
Pb(NO3)2 (aq) + Na2S (aq) NaNO3 (aq) + PbS (s)

67. Volume (L) of
Pb(NO3)2
solution
Volume (L)
of Na2S
solution
Multiply by
M (mol/L)
Multiply by
M (mol/L)
Amount (mol)
of Pb(NO3)2
Amount (mol)
of Na2S
Molar Ratio
Molar Ratio
Amount (mol)
of PbS
Amount (mol)
of PbS
Choose the lower number
of PbS and multiply by
M (g/mol)
Mass (g) of PbS

68. Volume (L) of
Pb(NO3)2
solution
Volume (L)
of Na2S
solution
Multiply by
M (mol/L)
Multiply by
M (mol/L)
Amount (mol)
of Pb(NO3)2
Divide by
equation
coefficient
Amount (mol)
of Na2S
Divide by
equation
coefficient
Smallest
Molar Ratio
Amount (mol)
of PbS
Mass (g) of PbS

69. Solving Limiting Reactant Problems in
Solution - Precipitation Problem - II
Moles Pb(NO3)2 = V x M = L x ( Mol/L) = =
Moles Na2S = V x M = L x (0.095 Mol/L) =

70. Solving Limiting Reactant Problems in
Solution - Precipitation Problem - II
Moles Pb(NO3)2 = V x M = L x ( Mol/L) =
= Mol Pb+2
Moles Na2S = V x M = L x (0.095 Mol/L) = mol S -2
Therefore Lead Nitrate is the Limiting Reactant!
Calculation of product yield:

71. Solving Limiting Reactant Problems in
Solution - Precipitation Problem - II
Moles Pb(NO3)2 = V x M = L x ( Mol/L) =
= Mol Pb+2
Moles Na2S = V x M = L x (0.095 Mol/L) = mol S -2
Therefore Lead Nitrate is the Limiting Reactant!
Calculation of product yield:
1 mol PbS
1 mol Pb(NO3)2
Moles PbS = Mol Pb+2x = Mol Pb+2

72. Solving Limiting Reactant Problems in
Solution - Precipitation Problem - II
Moles Pb(NO3)2 = V x M = L x ( Mol/L) =
= Mol Pb+2
Moles Na2S = V x M = L x (0.095 Mol/L) = mol S -2
Therefore Lead Nitrate is the Limiting Reactant!
Calculation of product yield:
1 mol PbS
1 mol Pb(NO3)2
Moles PbS = Mol Pb+2x = Mol Pb+2
Mol Pb+2 = Mol PbS
Mol PbS x = 2.89 g PbS
239.3 g PbS
1 Mol PbS

73. Quantitative Analysis
The determination of the amount of a substance or species present in a material 73

74. Gravimetric Analysis
A type of quantitative analysis in which the amount of a species in a material is determined by converting the species to a product that can be isolated completely and weighed 74

75. The figure on the right shows the reaction of Ba(NO3)2 with K2CrO4 forming the yellow BaCrO4 precipitate. 75

76. The BaCrO4 precipitate is being filtered in the figure on the right
The BaCrO4 precipitate is being filtered in the figure on the right. It can then be dried and weighed. 76

77. A soluble silver compound was analyzed for the percentage of silver by adding sodium chloride solution to precipitate the silver ion as silver chloride. If g of silver compound gave g of silver chloride, what is the mass percent of silver in the compound? 77

78. Molar mass of silver chloride (AgCl) = 143.32 g
= g Ag in the compound
= 85.03% Ag 78

80. Other Resources
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81. Other Resources
Visit the student website at college.hmco.com/pic/ebbing9e 81