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Franklin Ow

Franklin Ow

Stoichiometry II

Slide Duration:

Table of Contents

I. Basic Concepts & Measurement of Chemistry
Basic Concepts of Chemistry

16m 26s

Intro
0:00
Lesson Overview
0:07
Introduction
0:56
What is Chemistry?
0:57
What is Matter?
1:16
Solids
1:43
General Characteristics
1:44
Particulate-level Drawing of Solids
2:34
Liquids
3:39
General Characteristics of Liquids
3:40
Particulate-level Drawing of Liquids
3:55
Gases
4:23
General Characteristics of Gases
4:24
Particulate-level Drawing Gases
5:05
Classification of Matter
5:27
Classification of Matter
5:26
Pure Substances
5:54
Pure Substances
5:55
Mixtures
7:06
Definition of Mixtures
7:07
Homogeneous Mixtures
7:11
Heterogeneous Mixtures
7:52
Physical and Chemical Changes/Properties
8:18
Physical Changes Retain Chemical Composition
8:19
Chemical Changes Alter Chemical Composition
9:32
Physical and Chemical Changes/Properties, cont'd
10:55
Physical Properties
10:56
Chemical Properties
11:42
Sample Problem 1: Chemical & Physical Change
12:22
Sample Problem 2: Element, Compound, or Mixture?
13:52
Sample Problem 3: Classify Each of the Following Properties as chemical or Physical
15:03
Tools in Quantitative Chemistry

29m 22s

Intro
0:00
Lesson Overview
0:07
Units of Measurement
1:23
The International System of Units (SI): Mass, Length, and Volume
1:39
Percent Error
2:17
Percent Error
2:18
Example: Calculate the Percent Error
2:56
Standard Deviation
3:48
Standard Deviation Formula
3:49
Standard Deviation cont'd
4:42
Example: Calculate Your Standard Deviation
4:43
Precisions vs. Accuracy
6:25
Precision
6:26
Accuracy
7:01
Significant Figures and Uncertainty
7:50
Consider the Following (2) Rulers
7:51
Consider the Following Graduated Cylinder
11:30
Identifying Significant Figures
12:43
The Rules of Sig Figs Overview
12:44
The Rules for Sig Figs: All Nonzero Digits Are Significant
13:21
The Rules for Sig Figs: A Zero is Significant When It is In-Between Nonzero Digits
13:28
The Rules for Sig Figs: A Zero is Significant When at the End of a Decimal Number
14:02
The Rules for Sig Figs: A Zero is not significant When Starting a Decimal Number
14:27
Using Sig Figs in Calculations
15:03
Using Sig Figs for Multiplication and Division
15:04
Using Sig Figs for Addition and Subtraction
15:48
Using Sig Figs for Mixed Operations
16:11
Dimensional Analysis
16:20
Dimensional Analysis Overview
16:21
General Format for Dimensional Analysis
16:39
Example: How Many Miles are in 17 Laps?
17:17
Example: How Many Grams are in 1.22 Pounds?
18:40
Dimensional Analysis cont'd
19:43
Example: How Much is Spent on Diapers in One Week?
19:44
Dimensional Analysis cont'd
21:03
SI Prefixes
21:04
Dimensional Analysis cont'd
22:03
500 mg → ? kg
22:04
34.1 cm → ? um
24:03
Summary
25:11
Sample Problem 1: Dimensional Analysis
26:09
II. Atoms, Molecules, and Ions
Atoms, Molecules, and Ions

52m 18s

Intro
0:00
Lesson Overview
0:08
Introduction to Atomic Structure
1:03
Introduction to Atomic Structure
1:04
Plum Pudding Model
1:26
Introduction to Atomic Structure Cont'd
2:07
John Dalton's Atomic Theory: Number 1
2:22
John Dalton's Atomic Theory: Number 2
2:50
John Dalton's Atomic Theory: Number 3
3:07
John Dalton's Atomic Theory: Number 4
3:30
John Dalton's Atomic Theory: Number 5
3:58
Introduction to Atomic Structure Cont'd
5:21
Ernest Rutherford's Gold Foil Experiment
5:22
Introduction to Atomic Structure Cont'd
7:42
Implications of the Gold Foil Experiment
7:43
Relative Masses and Charges
8:18
Isotopes
9:02
Isotopes
9:03
Introduction to The Periodic Table
12:17
The Periodic Table of the Elements
12:18
Periodic Table, cont'd
13:56
Metals
13:57
Nonmetals
14:25
Semimetals
14:51
Periodic Table, cont'd
15:57
Group I: The Alkali Metals
15:58
Group II: The Alkali Earth Metals
16:25
Group VII: The Halogens
16:40
Group VIII: The Noble Gases
17:08
Ionic Compounds: Formulas, Names, Props.
17:35
Common Polyatomic Ions
17:36
Predicting Ionic Charge for Main Group Elements
18:52
Ionic Compounds: Formulas, Names, Props.
20:36
Naming Ionic Compounds: Rule 1
20:51
Naming Ionic Compounds: Rule 2
21:22
Naming Ionic Compounds: Rule 3
21:50
Naming Ionic Compounds: Rule 4
22:22
Ionic Compounds: Formulas, Names, Props.
22:50
Naming Ionic Compounds Example: Al₂O₃
22:51
Naming Ionic Compounds Example: FeCl₃
23:21
Naming Ionic Compounds Example: CuI₂ 3H₂O
24:00
Naming Ionic Compounds Example: Barium Phosphide
24:40
Naming Ionic Compounds Example: Ammonium Phosphate
25:55
Molecular Compounds: Formulas and Names
26:42
Molecular Compounds: Formulas and Names
26:43
The Mole
28:10
The Mole is 'A Chemist's Dozen'
28:11
It is a Central Unit, Connecting the Following Quantities
30:01
The Mole, cont'd
32:07
Atomic Masses
32:08
Example: How Many Moles are in 25.7 Grams of Sodium?
32:28
Example: How Many Atoms are in 1.2 Moles of Carbon?
33:17
The Mole, cont'd
34:25
Example: What is the Molar Mass of Carbon Dioxide?
34:26
Example: How Many Grams are in 1.2 Moles of Carbon Dioxide?
25:46
Percentage Composition
36:43
Example: How Many Grams of Carbon Contained in 65.1 Grams of Carbon Dioxide?
36:44
Empirical and Molecular Formulas
39:19
Empirical Formulas
39:20
Empirical Formula & Elemental Analysis
40:21
Empirical and Molecular Formulas, cont'd
41:24
Example: Determine Both the Empirical and Molecular Formulas - Step 1
41:25
Example: Determine Both the Empirical and Molecular Formulas - Step 2
43:18
Summary
46:22
Sample Problem 1: Determine the Empirical Formula of Lithium Fluoride
47:10
Sample Problem 2: How Many Atoms of Carbon are Present in 2.67 kg of C₆H₆?
49:21
III. Chemical Reactions
Chemical Reactions

43m 24s

Intro
0:00
Lesson Overview
0:06
The Law of Conservation of Mass and Balancing Chemical Reactions
1:49
The Law of Conservation of Mass
1:50
Balancing Chemical Reactions
2:50
Balancing Chemical Reactions Cont'd
3:40
Balance: N₂ + H₂ → NH₃
3:41
Balance: CH₄ + O₂ → CO₂ + H₂O
7:20
Balancing Chemical Reactions Cont'd
9:49
Balance: C₂H₆ + O₂ → CO₂ + H₂O
9:50
Intro to Chemical Equilibrium
15:32
When an Ionic Compound Full Dissociates
15:33
When an Ionic Compound Incompletely Dissociates
16:14
Dynamic Equilibrium
17:12
Electrolytes and Nonelectrolytes
18:03
Electrolytes
18:04
Strong Electrolytes and Weak Electrolytes
18:55
Nonelectrolytes
19:23
Predicting the Product(s) of an Aqueous Reaction
20:02
Single-replacement
20:03
Example: Li (s) + CuCl₂ (aq) → 2 LiCl (aq) + Cu (s)
21:03
Example: Cu (s) + LiCl (aq) → NR
21:23
Example: Zn (s) + 2HCl (aq) → ZnCl₂ (aq) + H₂ (g)
22:32
Predicting the Product(s) of an Aqueous Reaction
23:37
Double-replacement
23:38
Net-ionic Equation
25:29
Predicting the Product(s) of an Aqueous Reaction
26:12
Solubility Rules for Ionic Compounds
26:13
Predicting the Product(s) of an Aqueous Reaction
28:10
Neutralization Reactions
28:11
Example: HCl (aq) + NaOH (aq) → ?
28:37
Example: H₂SO₄ (aq) + KOH (aq) → ?
29:25
Predicting the Product(s) of an Aqueous Reaction
30:20
Certain Aqueous Reactions can Produce Unstable Compounds
30:21
Example 1
30:52
Example 2
32:16
Example 3
32:54
Summary
33:54
Sample Problem 1
34:55
ZnCO₃ (aq) + H₂SO₄ (aq) → ?
35:09
NH₄Br (aq) + Pb(C₂H₃O₂)₂ (aq) → ?
36:02
KNO₃ (aq) + CuCl₂ (aq) → ?
37:07
Li₂SO₄ (aq) + AgNO₃ (aq) → ?
37:52
Sample Problem 2
39:09
Question 1
39:10
Question 2
40:36
Question 3
41:47
Chemical Reactions II

55m 40s

Intro
0:00
Lesson Overview
0:10
Arrhenius Definition
1:15
Arrhenius Acids
1:16
Arrhenius Bases
3:20
The Bronsted-Lowry Definition
4:48
Acids Dissolve In Water and Donate a Proton to Water: Example 1
4:49
Acids Dissolve In Water and Donate a Proton to Water: Example 2
6:54
Monoprotic Acids & Polyprotic Acids
7:58
Strong Acids
11:30
Bases Dissolve In Water and Accept a Proton From Water
12:41
Strong Bases
16:36
The Autoionization of Water
17:42
Amphiprotic
17:43
Water Reacts With Itself
18:24
Oxides of Metals and Nonmetals
20:08
Oxides of Metals and Nonmetals Overview
20:09
Oxides of Nonmetals: Acidic Oxides
21:23
Oxides of Metals: Basic Oxides
24:08
Oxidation-Reduction (Redox) Reactions
25:34
Redox Reaction Overview
25:35
Oxidizing and Reducing Agents
27:02
Redox Reaction: Transfer of Electrons
27:54
Oxidation-Reduction Reactions Cont'd
29:55
Oxidation Number Overview
29:56
Oxidation Number of Homonuclear Species
31:17
Oxidation Number of Monatomic Ions
32:58
Oxidation Number of Fluorine
33:27
Oxidation Number of Oxygen
34:00
Oxidation Number of Chlorine, Bromine, and Iodine
35:07
Oxidation Number of Hydrogen
35:30
Net Sum of All Oxidation Numbers In a Compound
36:21
Oxidation-Reduction Reactions Cont'd
38:19
Let's Practice Assigning Oxidation Number
38:20
Now Let's Apply This to a Chemical Reaction
41:07
Summary
44:19
Sample Problems
45:29
Sample Problem 1
45:30
Sample Problem 2: Determine the Oxidizing and Reducing Agents
48:48
Sample Problem 3: Determine the Oxidizing and Reducing Agents
50:43
IV. Stoichiometry
Stoichiometry I

42m 10s

Intro
0:00
Lesson Overview
0:23
Mole to Mole Ratios
1:32
Example 1: In 1 Mole of H₂O, How Many Moles Are There of Each Element?
1:53
Example 2: In 2.6 Moles of Water, How Many Moles Are There of Each Element?
2:24
Mole to Mole Ratios Cont'd
5:13
Balanced Chemical Reaction
5:14
Mole to Mole Ratios Cont'd
7:25
Example 3: How Many Moles of Ammonia Can Form If you Have 3.1 Moles of H₂?
7:26
Example 4: How Many Moles of Hydrogen Gas Are Required to React With 6.4 Moles of Nitrogen Gas?
9:08
Mass to mass Conversion
11:06
Mass to mass Conversion
11:07
Example 5: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂?
12:37
Example 6: How Many Grams of Hydrogen Gas Are Required to React With 6.4 Grams of Nitrogen Gas?
15:34
Example 7: How Man Milligrams of Ammonia Can Form If You Have 1.2 kg of H₂?
17:29
Limiting Reactants, Percent Yields
20:42
Limiting Reactants, Percent Yields
20:43
Example 8: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂ and 3.1 Grams of N₂
22:25
Percent Yield
25:30
Example 9: How Many Grams of The Excess Reactant Remains?
26:37
Summary
29:34
Sample Problem 1: How Many Grams of Carbon Are In 2.2 Kilograms of Carbon Dioxide?
30:47
Sample Problem 2: How Many Milligrams of Carbon Dioxide Can Form From 23.1 Kg of CH₄(g)?
33:06
Sample Problem 3: Part 1
36:10
Sample Problem 3: Part 2 - What Amount Of The Excess Reactant Will Remain?
40:53
Stoichiometry II

42m 38s

Intro
0:00
Lesson Overview
0:10
Molarity
1:14
Solute and Solvent
1:15
Molarity
2:01
Molarity Cont'd
2:59
Example 1: How Many Grams of KBr are Needed to Make 350 mL of a 0.67 M KBr Solution?
3:00
Example 2: How Many Moles of KBr are in 350 mL of a 0.67 M KBr Solution?
5:44
Example 3: What Volume of a 0.67 M KBr Solution Contains 250 mg of KBr?
7:46
Dilutions
10:01
Dilution: M₁V₂=M₁V₂
10:02
Example 5: Explain How to Make 250 mL of a 0.67 M KBr Solution Starting From a 1.2M Stock Solution
12:04
Stoichiometry and Double-Displacement Precipitation Reactions
14:41
Example 6: How Many grams of PbCl₂ Can Form From 250 mL of 0.32 M NaCl?
15:38
Stoichiometry and Double-Displacement Precipitation Reactions
18:05
Example 7: How Many grams of PbCl₂ Can Form When 250 mL of 0.32 M NaCl and 150 mL of 0.45 Pb(NO₃)₂ Mix?
18:06
Stoichiometry and Neutralization Reactions
21:01
Example 8: How Many Grams of NaOh are Required to Neutralize 4.5 Grams of HCl?
21:02
Stoichiometry and Neutralization Reactions
23:03
Example 9: How Many mL of 0.45 M NaOH are Required to Neutralize 250 mL of 0.89 M HCl?
23:04
Stoichiometry and Acid-Base Standardization
25:28
Introduction to Titration & Standardization
25:30
Acid-Base Titration
26:12
The Analyte & Titrant
26:24
The Experimental Setup
26:49
The Experimental Setup
26:50
Stoichiometry and Acid-Base Standardization
28:38
Example 9: Determine the Concentration of the Analyte
28:39
Summary
32:46
Sample Problem 1: Stoichiometry & Neutralization
35:24
Sample Problem 2: Stoichiometry
37:50
V. Thermochemistry
Energy & Chemical Reactions

55m 28s

Intro
0:00
Lesson Overview
0:14
Introduction
1:22
Recall: Chemistry
1:23
Energy Can Be Expressed In Different Units
1:57
The First Law of Thermodynamics
2:43
Internal Energy
2:44
The First Law of Thermodynamics Cont'd
6:14
Ways to Transfer Internal Energy
6:15
Work Energy
8:13
Heat Energy
8:34
∆U = q + w
8:44
Calculating ∆U, Q, and W
8:58
Changes In Both Volume and Temperature of a System
8:59
Calculating ∆U, Q, and W Cont'd
11:01
The Work Equation
11:02
Example 1: Calculate ∆U For The Burning Fuel
11:45
Calculating ∆U, Q, and W Cont'd
14:09
The Heat Equation
14:10
Calculating ∆U, Q, and W Cont'd
16:03
Example 2: Calculate The Final Temperature
16:04
Constant-Volume Calorimetry
18:05
Bomb Calorimeter
18:06
The Effect of Constant Volume On The Equation For Internal Energy
22:11
Example 3: Calculate ∆U
23:12
Constant-Pressure Conditions
26:05
Constant-Pressure Conditions
26:06
Calculating Enthalpy: Phase Changes
27:29
Melting, Vaporization, and Sublimation
27:30
Freezing, Condensation and Deposition
28:25
Enthalpy Values For Phase Changes
28:40
Example 4: How Much Energy In The Form of heat is Required to Melt 1.36 Grams of Ice?
29:40
Calculating Enthalpy: Heats of Reaction
31:22
Example 5: Calculate The Heat In kJ Associated With The Complete Reaction of 155 g NH₃
31:23
Using Standard Enthalpies of Formation
33:53
Standard Enthalpies of Formation
33:54
Using Standard Enthalpies of Formation
36:12
Example 6: Calculate The Standard Enthalpies of Formation For The Following Reaction
36:13
Enthalpy From a Series of Reactions
39:58
Hess's Law
39:59
Coffee-Cup Calorimetry
42:43
Coffee-Cup Calorimetry
42:44
Example 7: Calculate ∆H° of Reaction
45:10
Summary
47:12
Sample Problem 1
48:58
Sample Problem 2
51:24
VI. Quantum Theory of Atoms
Structure of Atoms

42m 33s

Intro
0:00
Lesson Overview
0:07
Introduction
1:01
Rutherford's Gold Foil Experiment
1:02
Electromagnetic Radiation
2:31
Radiation
2:32
Three Parameters: Energy, Frequency, and Wavelength
2:52
Electromagnetic Radiation
5:18
The Electromagnetic Spectrum
5:19
Atomic Spectroscopy and The Bohr Model
7:46
Wavelengths of Light
7:47
Atomic Spectroscopy Cont'd
9:45
The Bohr Model
9:46
Atomic Spectroscopy Cont'd
12:21
The Balmer Series
12:22
Rydberg Equation For Predicting The Wavelengths of Light
13:04
The Wave Nature of Matter
15:11
The Wave Nature of Matter
15:12
The Wave Nature of Matter
19:10
New School of Thought
19:11
Einstein: Energy
19:49
Hertz and Planck: Photoelectric Effect
20:16
de Broglie: Wavelength of a Moving Particle
21:14
Quantum Mechanics and The Atom
22:15
Heisenberg: Uncertainty Principle
22:16
Schrodinger: Wavefunctions
23:08
Quantum Mechanics and The Atom
24:02
Principle Quantum Number
24:03
Angular Momentum Quantum Number
25:06
Magnetic Quantum Number
26:27
Spin Quantum Number
28:42
The Shapes of Atomic Orbitals
29:15
Radial Wave Function
29:16
Probability Distribution Function
32:08
The Shapes of Atomic Orbitals
34:02
3-Dimensional Space of Wavefunctions
34:03
Summary
35:57
Sample Problem 1
37:07
Sample Problem 2
40:23
VII. Electron Configurations and Periodicity
Periodic Trends

38m 50s

Intro
0:00
Lesson Overview
0:09
Introduction
0:36
Electron Configuration of Atoms
1:33
Electron Configuration & Atom's Electrons
1:34
Electron Configuration Format
1:56
Electron Configuration of Atoms Cont'd
3:01
Aufbau Principle
3:02
Electron Configuration of Atoms Cont'd
6:53
Electron Configuration Format 1: Li, O, and Cl
6:56
Electron Configuration Format 2: Li, O, and Cl
9:11
Electron Configuration of Atoms Cont'd
12:48
Orbital Box Diagrams
12:49
Pauli Exclusion Principle
13:11
Hund's Rule
13:36
Electron Configuration of Atoms Cont'd
17:35
Exceptions to The Aufbau Principle: Cr
17:36
Exceptions to The Aufbau Principle: Cu
18:15
Electron Configuration of Atoms Cont'd
20:22
Electron Configuration of Monatomic Ions: Al
20:23
Electron Configuration of Monatomic Ions: Al³⁺
20:46
Electron Configuration of Monatomic Ions: Cl
21:57
Electron Configuration of Monatomic Ions: Cl¹⁻
22:09
Electron Configuration Cont'd
24:31
Paramagnetism
24:32
Diamagnetism
25:00
Atomic Radii
26:08
Atomic Radii
26:09
In a Column of the Periodic Table
26:25
In a Row of the Periodic Table
26:46
Ionic Radii
27:30
Ionic Radii
27:31
Anions
27:42
Cations
27:57
Isoelectronic Species
28:12
Ionization Energy
29:00
Ionization Energy
29:01
Electron Affinity
31:37
Electron Affinity
31:37
Summary
33:43
Sample Problem 1: Ground State Configuration and Orbital Box Diagram
34:21
Fe
34:48
P
35:32
Sample Problem 2
36:38
Which Has The Larger Ionization Energy: Na or Li?
36:39
Which Has The Larger Atomic Size: O or N ?
37:23
Which Has The Larger Atomic Size: O²⁻ or N³⁻ ?
38:00
VIII. Molecular Geometry & Bonding Theory
Bonding & Molecular Structure

52m 39s

Intro
0:00
Lesson Overview
0:08
Introduction
1:10
Types of Chemical Bonds
1:53
Ionic Bond
1:54
Molecular Bond
2:42
Electronegativity and Bond Polarity
3:26
Electronegativity (EN)
3:27
Periodic Trend
4:36
Electronegativity and Bond Polarity Cont'd
6:04
Bond Polarity: Polar Covalent Bond
6:05
Bond Polarity: Nonpolar Covalent Bond
8:53
Lewis Electron Dot Structure of Atoms
9:48
Lewis Electron Dot Structure of Atoms
9:49
Lewis Structures of Polyatomic Species
12:51
Single Bonds
12:52
Double Bonds
13:28
Nonbonding Electrons
13:59
Lewis Structures of Polyatomic Species Cont'd
14:45
Drawing Lewis Structures: Step 1
14:48
Drawing Lewis Structures: Step 2
15:16
Drawing Lewis Structures: Step 3
15:52
Drawing Lewis Structures: Step 4
17:31
Drawing Lewis Structures: Step 5
19:08
Drawing Lewis Structure Example: Carbonate
19:33
Resonance and Formal Charges (FC)
24:06
Resonance Structures
24:07
Formal Charge
25:20
Resonance and Formal Charges Cont'd
27:46
More On Formal Charge
27:47
Resonance and Formal Charges Cont'd
28:21
Good Resonance Structures
28:22
VSEPR Theory
31:08
VSEPR Theory Continue
31:09
VSEPR Theory Cont'd
32:53
VSEPR Geometries
32:54
Steric Number
33:04
Basic Geometry
33:50
Molecular Geometry
35:50
Molecular Polarity
37:51
Steps In Determining Molecular Polarity
37:52
Example 1: Polar
38:47
Example 2: Nonpolar
39:10
Example 3: Polar
39:36
Example 4: Polar
40:08
Bond Properties: Order, Length, and Energy
40:38
Bond Order
40:39
Bond Length
41:21
Bond Energy
41:55
Summary
43:09
Sample Problem 1
43:42
XeO₃
44:03
I₃⁻
47:02
SF₅
49:16
Advanced Bonding Theories

1h 11m 41s

Intro
0:00
Lesson Overview
0:09
Introduction
0:38
Valence Bond Theory
3:07
Valence Bond Theory
3:08
spᶟ Hybridized Carbon Atom
4:19
Valence Bond Theory Cont'd
6:24
spᶟ Hybridized
6:25
Hybrid Orbitals For Water
7:26
Valence Bond Theory Cont'd (spᶟ)
11:53
Example 1: NH₃
11:54
Valence Bond Theory Cont'd (sp²)
14:48
sp² Hybridization
14:49
Example 2: BF₃
16:44
Valence Bond Theory Cont'd (sp)
22:44
sp Hybridization
22:46
Example 3: HCN
23:38
Valence Bond Theory Cont'd (sp³d and sp³d²)
27:36
Valence Bond Theory: sp³d and sp³d²
27:37
Molecular Orbital Theory
29:10
Valence Bond Theory Doesn't Always Account For a Molecule's Magnetic Behavior
29:11
Molecular Orbital Theory Cont'd
30:37
Molecular Orbital Theory
30:38
Wavefunctions
31:04
How s-orbitals Can Interact
32:23
Bonding Nature of p-orbitals: Head-on
35:34
Bonding Nature of p-orbitals: Parallel
39:04
Interaction Between s and p-orbital
40:45
Molecular Orbital Diagram For Homonuclear Diatomics: H₂
42:21
Molecular Orbital Diagram For Homonuclear Diatomics: He₂
45:23
Molecular Orbital Diagram For Homonuclear Diatomic: Li₂
46:39
Molecular Orbital Diagram For Homonuclear Diatomic: Li₂⁺
47:42
Molecular Orbital Diagram For Homonuclear Diatomic: B₂
48:57
Molecular Orbital Diagram For Homonuclear Diatomic: N₂
54:04
Molecular Orbital Diagram: Molecular Oxygen
55:57
Molecular Orbital Diagram For Heteronuclear Diatomics: Hydrochloric Acid
1:02:16
Sample Problem 1: Determine the Atomic Hybridization
1:07:20
XeO₃
1:07:21
SF₆
1:07:49
I₃⁻
1:08:20
Sample Problem 2
1:09:04
IX. Gases, Solids, & Liquids
Gases

35m 6s

Intro
0:00
Lesson Overview
0:07
The Kinetic Molecular Theory of Gases
1:23
The Kinetic Molecular Theory of Gases
1:24
Parameters To Characterize Gases
3:35
Parameters To Characterize Gases: Pressure
3:37
Interpreting Pressure On a Particulate Level
4:43
Parameters Cont'd
6:08
Units For Expressing Pressure: Psi, Pascal
6:19
Units For Expressing Pressure: mm Hg
6:42
Units For Expressing Pressure: atm
6:58
Units For Expressing Pressure: torr
7:24
Parameters Cont'd
8:09
Parameters To Characterize Gases: Volume
8:10
Common Units of Volume
9:00
Parameters Cont'd
9:11
Parameters To Characterize Gases: Temperature
9:12
Particulate Level
9:36
Parameters To Characterize Gases: Moles
10:24
The Simple Gas Laws
10:43
Gas Laws Are Only Valid For…
10:44
Charles' Law
11:24
The Simple Gas Laws
13:13
Boyle's Law
13:14
The Simple Gas Laws
15:28
Gay-Lussac's Law
15:29
The Simple Gas Laws
17:11
Avogadro's Law
17:12
The Ideal Gas Law
18:43
The Ideal Gas Law: PV = nRT
18:44
Applications of the Ideal Gas Law
20:12
Standard Temperature and Pressure for Gases
20:13
Applications of the Ideal Gas Law
21:43
Ideal Gas Law & Gas Density
21:44
Gas Pressures and Partial Pressures
23:18
Dalton's Law of Partial Pressures
23:19
Gas Stoichiometry
24:15
Stoichiometry Problems Involving Gases
24:16
Using The Ideal Gas Law to Get to Moles
25:16
Using Molar Volume to Get to Moles
25:39
Gas Stoichiometry Cont'd
26:03
Example 1: How Many Liters of O₂ at STP are Needed to Form 10.5 g of Water Vapor?
26:04
Summary
28:33
Sample Problem 1: Calculate the Molar Mass of the Gas
29:28
Sample Problem 2: What Mass of Ag₂O is Required to Form 3888 mL of O₂ Gas When Measured at 734 mm Hg and 25°C?
31:59
Intermolecular Forces & Liquids

33m 47s

Intro
0:00
Lesson Overview
0:10
Introduction
0:46
Intermolecular Forces (IMF)
0:47
Intermolecular Forces of Polar Molecules
1:32
Ion-dipole Forces
1:33
Example: Salt Dissolved in Water
1:50
Coulomb's Law & the Force of Attraction Between Ions and/or Dipoles
3:06
IMF of Polar Molecules cont'd
4:36
Enthalpy of Solvation or Enthalpy of Hydration
4:37
IMF of Polar Molecules cont'd
6:01
Dipole-dipole Forces
6:02
IMF of Polar Molecules cont'd
7:22
Hydrogen Bonding
7:23
Example: Hydrogen Bonding of Water
8:06
IMF of Nonpolar Molecules
9:37
Dipole-induced Dipole Attraction
9:38
IMF of Nonpolar Molecules cont'd
11:34
Induced Dipole Attraction, London Dispersion Forces, or Vand der Waals Forces
11:35
Polarizability
13:46
IMF of Nonpolar Molecules cont'd
14:26
Intermolecular Forces (IMF) and Polarizability
14:31
Properties of Liquids
16:48
Standard Molar Enthalpy of Vaporization
16:49
Trends in Boiling Points of Representative Liquids: H₂O vs. H₂S
17:43
Properties of Liquids cont'd
18:36
Aliphatic Hydrocarbons
18:37
Branched Hydrocarbons
20:52
Properties of Liquids cont'd
22:10
Vapor Pressure
22:11
The Clausius-Clapeyron Equation
24:30
Properties of Liquids cont'd
25:52
Boiling Point
25:53
Properties of Liquids cont'd
27:07
Surface Tension
27:08
Viscosity
28:06
Summary
29:04
Sample Problem 1: Determine Which of the Following Liquids Will Have the Lower Vapor Pressure
30:21
Sample Problem 2: Determine Which of the Following Liquids Will Have the Largest Standard Molar Enthalpy of Vaporization
31:37
The Chemistry of Solids

25m 13s

Intro
0:00
Lesson Overview
0:07
Introduction
0:46
General Characteristics
0:47
Particulate-level Drawing
1:09
The Basic Structure of Solids: Crystal Lattices
1:37
The Unit Cell Defined
1:38
Primitive Cubic
2:50
Crystal Lattices cont'd
3:58
Body-centered Cubic
3:59
Face-centered Cubic
5:02
Lattice Enthalpy and Trends
6:27
Introduction to Lattice Enthalpy
6:28
Equation to Calculate Lattice Enthalpy
7:21
Different Types of Crystalline Solids
9:35
Molecular Solids
9:36
Network Solids
10:25
Phase Changes Involving Solids
11:03
Melting & Thermodynamic Value
11:04
Freezing & Thermodynamic Value
11:49
Phase Changes cont'd
12:40
Sublimation & Thermodynamic Value
12:41
Depositions & Thermodynamic Value
13:13
Phase Diagrams
13:40
Introduction to Phase Diagrams
13:41
Phase Diagram of H₂O: Melting Point
14:12
Phase Diagram of H₂O: Normal Boiling Point
14:50
Phase Diagram of H₂O: Sublimation Point
15:02
Phase Diagram of H₂O: Point C ( Supercritical Point)
15:32
Phase Diagrams cont'd
16:31
Phase Diagram of Dry Ice
16:32
Summary
18:15
Sample Problem 1, Part A: Of the Group I Fluorides, Which Should Have the Highest Lattice Enthalpy?
19:01
Sample Problem 1, Part B: Of the Lithium Halides, Which Should Have the Lowest Lattice Enthalpy?
19:54
Sample Problem 2: How Many Joules of Energy is Required to Melt 546 mg of Ice at Standard Pressure?
20:55
Sample Problem 3: Phase Diagram of Helium
22:42
X. Solutions, Rates of Reaction, & Equilibrium
Solutions & Their Behavior

38m 6s

Intro
0:00
Lesson Overview
0:10
Units of Concentration
1:40
Molarity
1:41
Molality
3:30
Weight Percent
4:26
ppm
5:16
Like Dissolves Like
6:28
Like Dissolves Like
6:29
Factors Affecting Solubility
9:35
The Effect of Pressure: Henry's Law
9:36
The Effect of Temperature on Gas Solubility
12:16
The Effect of Temperature on Solid Solubility
14:28
Colligative Properties
16:48
Colligative Properties
16:49
Changes in Vapor Pressure: Raoult's Law
17:19
Colligative Properties cont'd
19:53
Boiling Point Elevation and Freezing Point Depression
19:54
Colligative Properties cont'd
26:13
Definition of Osmosis
26:14
Osmotic Pressure Example
27:11
Summary
31:11
Sample Problem 1: Calculating Vapor Pressure
32:53
Sample Problem 2: Calculating Molality
36:29
Chemical Kinetics

37m 45s

Intro
0:00
Lesson Overview
0:06
Introduction
1:09
Chemical Kinetics and the Rate of a Reaction
1:10
Factors Influencing Rate
1:19
Introduction cont'd
2:27
How a Reaction Progresses Through Time
2:28
Rate of Change Equation
6:02
Rate Laws
7:06
Definition of Rate Laws
7:07
General Form of Rate Laws
7:37
Rate Laws cont'd
11:07
Rate Orders With Respect to Reactant and Concentration
11:08
Methods of Initial Rates
13:38
Methods of Initial Rates
13:39
Integrated Rate Laws
17:57
Integrated Rate Laws
17:58
Graphically Determine the Rate Constant k
18:52
Reaction Mechanisms
21:05
Step 1: Reversible
21:18
Step 2: Rate-limiting Step
21:44
Rate Law for the Reaction
23:28
Reaction Rates and Temperatures
26:16
Reaction Rates and Temperatures
26:17
The Arrhenius Equation
29:06
Catalysis
30:31
Catalyst
30:32
Summary
32:02
Sample Problem 1: Calculate the Rate Constant and the Time Required for the Reaction to be Completed
32:54
Sample Problem 2: Calculate the Energy of Activation and the Order of the Reaction
35:24
Principles of Chemical Equilibrium

34m 9s

Intro
0:00
Lesson Overview
0:08
Introduction
1:02
The Equilibrium Constant
3:08
The Equilibrium Constant
3:09
The Equilibrium Constant cont'd
5:50
The Equilibrium Concentration and Constant for Solutions
5:51
The Equilibrium Partial Pressure and Constant for Gases
7:01
Relationship of Kc and Kp
7:30
Heterogeneous Equilibria
8:23
Heterogeneous Equilibria
8:24
Manipulating K
9:57
First Way of Manipulating K
9:58
Second Way of Manipulating K
11:48
Manipulating K cont'd
12:31
Third Way of Manipulating K
12:32
The Reaction Quotient Q
14:42
The Reaction Quotient Q
14:43
Q > K
16:16
Q < K
16:30
Q = K
16:43
Le Chatlier's Principle
17:32
Restoring Equilibrium When It is Disturbed
17:33
Disturbing a Chemical System at Equilibrium
18:35
Problem-Solving with ICE Tables
19:05
Determining a Reaction's Equilibrium Constant With ICE Table
19:06
Problem-Solving with ICE Tables cont'd
21:03
Example 1: Calculate O₂(g) at Equilibrium
21:04
Problem-Solving with ICE Tables cont'd
22:53
Example 2: Calculate the Equilibrium Constant
22:54
Summary
25:24
Sample Problem 1: Calculate the Equilibrium Constant
27:59
Sample Problem 2: Calculate The Equilibrium Concentration
30:30
XI. Acids & Bases Chemistry
Acid-Base Chemistry

43m 44s

Intro
0:00
Lesson Overview
0:06
Introduction
0:55
Bronsted-Lowry Acid & Bronsted -Lowry Base
0:56
Water is an Amphiprotic Molecule
2:40
Water Reacting With Itself
2:58
Introduction cont'd
4:04
Strong Acids
4:05
Strong Bases
5:18
Introduction cont'd
6:16
Weak Acids and Bases
6:17
Quantifying Acid-Base Strength
7:35
The pH Scale
7:36
Quantifying Acid-Base Strength cont'd
9:55
The Acid-ionization Constant Ka and pKa
9:56
Quantifying Acid-Base Strength cont'd
12:13
Example: Calculate the pH of a 1.2M Solution of Acetic Acid
12:14
Quantifying Acid-Base Strength
15:06
Calculating the pH of Weak Base Solutions
15:07
Writing Out Acid-Base Equilibria
17:45
Writing Out Acid-Base Equilibria
17:46
Writing Out Acid-Base Equilibria cont'd
19:47
Consider the Following Equilibrium
19:48
Conjugate Base and Conjugate Acid
21:18
Salts Solutions
22:00
Salts That Produce Acidic Aqueous Solutions
22:01
Salts That Produce Basic Aqueous Solutions
23:15
Neutral Salt Solutions
24:05
Diprotic and Polyprotic Acids
24:44
Example: Calculate the pH of a 1.2 M Solution of H₂SO₃
24:43
Diprotic and Polyprotic Acids cont'd
27:18
Calculate the pH of a 1.2 M Solution of Na₂SO₃
27:19
Lewis Acids and Bases
29:13
Lewis Acids
29:14
Lewis Bases
30:10
Example: Lewis Acids and Bases
31:04
Molecular Structure and Acidity
32:03
The Effect of Charge
32:04
Within a Period/Row
33:07
Molecular Structure and Acidity cont'd
34:17
Within a Group/Column
34:18
Oxoacids
35:58
Molecular Structure and Acidity cont'd
37:54
Carboxylic Acids
37:55
Hydrated Metal Cations
39:23
Summary
40:39
Sample Problem 1: Calculate the pH of a 1.2 M Solution of NH₃
41:20
Sample Problem 2: Predict If The Following Slat Solutions are Acidic, Basic, or Neutral
42:37
Applications of Aqueous Equilibria

55m 26s

Intro
0:00
Lesson Overview
0:07
Calculating pH of an Acid-Base Mixture
0:53
Equilibria Involving Direct Reaction With Water
0:54
When a Bronsted-Lowry Acid and Base React
1:12
After Neutralization Occurs
2:05
Calculating pH of an Acid-Base Mixture cont'd
2:51
Example: Calculating pH of an Acid-Base Mixture, Step 1 - Neutralization
2:52
Example: Calculating pH of an Acid-Base Mixture, Step 2 - React With H₂O
5:24
Buffers
7:45
Introduction to Buffers
7:46
When Acid is Added to a Buffer
8:50
When Base is Added to a Buffer
9:54
Buffers cont'd
10:41
Calculating the pH
10:42
Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer
14:03
Buffers cont'd
14:10
Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer: Step 1 -Neutralization
14:11
Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer: Step 2- ICE Table
15:22
Buffer Preparation and Capacity
16:38
Example: Calculating the pH of a Buffer Solution
16:42
Effective Buffer
18:40
Acid-Base Titrations
19:33
Acid-Base Titrations: Basic Setup
19:34
Acid-Base Titrations cont'd
22:12
Example: Calculate the pH at the Equivalence Point When 0.250 L of 0.0350 M HClO is Titrated With 1.00 M KOH
22:13
Acid-Base Titrations cont'd
25:38
Titration Curve
25:39
Solubility Equilibria
33:07
Solubility of Salts
33:08
Solubility Product Constant: Ksp
34:14
Solubility Equilibria cont'd
34:58
Q < Ksp
34:59
Q > Ksp
35:34
Solubility Equilibria cont'd
36:03
Common-ion Effect
36:04
Example: Calculate the Solubility of PbCl₂ in 0.55 M NaCl
36:30
Solubility Equilibria cont'd
39:02
When a Solid Salt Contains the Conjugate of a Weak Acid
39:03
Temperature and Solubility
40:41
Complexation Equilibria
41:10
Complex Ion
41:11
Complex Ion Formation Constant: Kf
42:26
Summary
43:35
Sample Problem 1: Question
44:23
Sample Problem 1: Part a) Calculate the pH at the Beginning of the Titration
45:48
Sample Problem 1: Part b) Calculate the pH at the Midpoint or Half-way Point
48:04
Sample Problem 1: Part c) Calculate the pH at the Equivalence Point
48:32
Sample Problem 1: Part d) Calculate the pH After 27.50 mL of the Acid was Added
53:00
XII. Thermodynamics & Electrochemistry
Entropy & Free Energy

36m 13s

Intro
0:00
Lesson Overview
0:08
Introduction
0:53
Introduction to Entropy
1:37
Introduction to Entropy
1:38
Entropy and Heat Flow
6:31
Recall Thermodynamics
6:32
Entropy is a State Function
6:54
∆S and Heat Flow
7:28
Entropy and Heat Flow cont'd
8:18
Entropy and Heat Flow: Equations
8:19
Endothermic Processes: ∆S > 0
8:44
The Second Law of Thermodynamics
10:04
Total ∆S = ∆S of System + ∆S of Surrounding
10:05
Nature Favors Processes Where The Amount of Entropy Increases
10:22
The Third Law of Thermodynamics
11:55
The Third Law of Thermodynamics & Zero Entropy
11:56
Problem-Solving involving Entropy
12:36
Endothermic Process and ∆S
12:37
Exothermic Process and ∆S
13:19
Problem-Solving cont'd
13:46
Change in Physical States: From Solid to Liquid to Gas
13:47
Change in Physical States: All Gases
15:02
Problem-Solving cont'd
15:56
Calculating the ∆S for the System, Surrounding, and Total
15:57
Example: Calculating the Total ∆S
16:17
Problem-Solving cont'd
18:36
Problems Involving Standard Molar Entropies of Formation
18:37
Introduction to Gibb's Free Energy
20:09
Definition of Free Energy ∆G
20:10
Spontaneous Process and ∆G
20:19
Gibb's Free Energy cont'd
22:28
Standard Molar Free Energies of Formation
22:29
The Free Energies of Formation are Zero for All Compounds in the Standard State
22:42
Gibb's Free Energy cont'd
23:31
∆G° of the System = ∆H° of the System - T∆S° of the System
23:32
Predicting Spontaneous Reaction Based on the Sign of ∆G° of the System
24:24
Gibb's Free Energy cont'd
26:32
Effect of reactant and Product Concentration on the Sign of Free Energy
26:33
∆G° of Reaction = -RT ln K
27:18
Summary
28:12
Sample Problem 1: Calculate ∆S° of Reaction
28:48
Sample Problem 2: Calculate the Temperature at Which the Reaction Becomes Spontaneous
31:18
Sample Problem 3: Calculate Kp
33:47
Electrochemistry

41m 16s

Intro
0:00
Lesson Overview
0:08
Introduction
0:53
Redox Reactions
1:42
Oxidation-Reduction Reaction Overview
1:43
Redox Reactions cont'd
2:37
Which Reactant is Being Oxidized and Which is Being Reduced?
2:38
Redox Reactions cont'd
6:34
Balance Redox Reaction In Neutral Solutions
6:35
Redox Reactions cont'd
10:37
Balance Redox Reaction In Acidic and Basic Solutions: Step 1
10:38
Balance Redox Reaction In Acidic and Basic Solutions: Step 2 - Balance Each Half-Reaction
11:22
Redox Reactions cont'd
12:19
Balance Redox Reaction In Acidic and Basic Solutions: Step 2 - Balance Hydrogen
12:20
Redox Reactions cont'd
14:30
Balance Redox Reaction In Acidic and Basic Solutions: Step 3
14:34
Balance Redox Reaction In Acidic and Basic Solutions: Step 4
15:38
Voltaic Cells
17:01
Voltaic Cell or Galvanic Cell
17:02
Cell Notation
22:03
Electrochemical Potentials
25:22
Electrochemical Potentials
25:23
Electrochemical Potentials cont'd
26:07
Table of Standard Reduction Potentials
26:08
The Nernst Equation
30:41
The Nernst Equation
30:42
It Can Be Shown That At Equilibrium E =0.00
32:15
Gibb's Free Energy and Electrochemistry
32:46
Gibbs Free Energy is Relatively Small if the Potential is Relatively High
32:47
When E° is Very Large
33:39
Charge, Current and Time
33:56
A Battery Has Three Main Parameters
33:57
A Simple Equation Relates All of These Parameters
34:09
Summary
34:50
Sample Problem 1: Redox Reaction
35:26
Sample Problem 2: Battery
38:00
XIII. Transition Elements & Coordination Compounds
The Chemistry of The Transition Metals

39m 3s

Intro
0:00
Lesson Overview
0:11
Coordination Compounds
1:20
Coordination Compounds
1:21
Nomenclature of Coordination Compounds
2:48
Rule 1
3:01
Rule 2
3:12
Rule 3
4:07
Nomenclature cont'd
4:58
Rule 4
4:59
Rule 5
5:13
Rule 6
5:35
Rule 7
6:19
Rule 8
6:46
Nomenclature cont'd
7:39
Rule 9
7:40
Rule 10
7:45
Rule 11
8:00
Nomenclature of Coordination Compounds: NH₄[PtCl₃NH₃]
8:11
Nomenclature of Coordination Compounds: [Cr(NH₃)₄(OH)₂]Br
9:31
Structures of Coordination Compounds
10:54
Coordination Number or Steric Number
10:55
Commonly Observed Coordination Numbers and Geometries: 4
11:14
Commonly Observed Coordination Numbers and Geometries: 6
12:00
Isomers of Coordination Compounds
13:13
Isomers of Coordination Compounds
13:14
Geometrical Isomers of CN = 6 Include: ML₄L₂'
13:30
Geometrical Isomers of CN = 6 Include: ML₃L₃'
15:07
Isomers cont'd
17:00
Structural Isomers Overview
17:01
Structural Isomers: Ionization
18:06
Structural Isomers: Hydrate
19:25
Structural Isomers: Linkage
20:11
Structural Isomers: Coordination Isomers
21:05
Electronic Structure
22:25
Crystal Field Theory
22:26
Octahedral and Tetrahedral Field
22:54
Electronic Structure cont'd
25:43
Vanadium (II) Ion in an Octahedral Field
25:44
Chromium(III) Ion in an Octahedral Field
26:37
Electronic Structure cont'd
28:47
Strong-Field Ligands and Weak-Field Ligands
28:48
Implications of Electronic Structure
30:08
Compare the Magnetic Properties of: [Fe(OH₂)₆]²⁺ vs. [Fe(CN)₆]⁴⁻
30:09
Discussion on Color
31:57
Summary
34:41
Sample Problem 1: Name the Following Compound [Fe(OH)(OH₂)₅]Cl₂
35:08
Sample Problem 1: Name the Following Compound [Co(NH₃)₃(OH₂)₃]₂(SO₄)₃
36:24
Sample Problem 2: Change in Magnetic Properties
37:30
XIV. Nuclear Chemistry
Nuclear Chemistry

16m 39s

Intro
0:00
Lesson Overview
0:06
Introduction
0:40
Introduction to Nuclear Reactions
0:41
Types of Radioactive Decay
2:10
Alpha Decay
2:11
Beta Decay
3:27
Gamma Decay
4:40
Other Types of Particles of Varying Energy
5:40
Nuclear Equations
6:47
Nuclear Equations
6:48
Nuclear Decay
9:28
Nuclear Decay and the First-Order Kinetics
9:29
Summary
11:31
Sample Problem 1: Complete the Following Nuclear Equations
12:13
Sample Problem 2: How Old is the Rock?
14:21
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Lecture Comments (21)

0 answers

Post by Evan Wang on May 18, 2017

M1V1=M2V2 looks like that physics equation but with mass and velocity. Are the two based off of the same comment?

1 answer

Last reply by: Professor Franklin Ow
Thu May 28, 2015 12:28 PM

Post by Mikayla McNulty on May 13, 2015

For question #9 shouldn't 32.10 mL be 0.0321L not 0.321L?

1 answer

Last reply by: Professor Franklin Ow
Mon Nov 3, 2014 10:59 PM

Post by Saadman Elman on October 31, 2014

Diego Robledo already addressed it before but he didn't give the whole picture. There is a mistake in example no. 9 (Stochiometry and Acid base Standardization) The mistake is you said 0.011 moles of Mg(OH)2 is produced. If you do the calculation again you will see that the answer is actually 0.107 moles. Even though it is a minute mistake at the beginning it made a huge difference in the end. Because later on we had to find the molarity which is moles divided by liter. If you divide .107 divided by  .0250 you will get 4.3 Molarity and NOT 0.43 Molarity. You got .43

But overall, the lecture was awesome! I benefited from your lecture a lot. And i did great in the class. Sometimes even my professor took some notes from me when he was teaching ionization energy. He was really impressed. I can't thank you enough! You really made me change my major! I listened to all of your lectures!!

1 answer

Last reply by: Professor Franklin Ow
Mon Nov 3, 2014 10:59 PM

Post by Minky benedikt on October 23, 2014

Hi professor,
In example 5, can you please write the full equation to solve for V1?

1 answer

Last reply by: Professor Franklin Ow
Mon Oct 20, 2014 12:31 AM

Post by Joseph Patrick Balao on October 19, 2014

Hi Professor,
How do you calculate this problem
How many milliliters of 0.250 M hydrochloric acid are required to neutralize a solution containing 0.863 g barium hydroxide ?

1 answer

Last reply by: Professor Franklin Ow
Wed Jul 2, 2014 11:51 AM

Post by brandon joyner on July 1, 2014

Why is 1.2 M stock solution M1 and not M2

0 answers

Post by Saadman Elman on May 29, 2014

It's really helpful! Thanks a lot!

1 answer

Last reply by: Professor Franklin Ow
Sun Mar 16, 2014 12:26 AM

Post by Meredith Roach on March 9, 2014

On sample problem #7, how is it possible that the LR is NaCl when Pb(NO3)2 actually yields less product?  To calculate Theoretical Yield, we would have to use the reactant that produces the smallest amount of product, no?

2 answers

Last reply by: Saadman Elman
Fri Oct 31, 2014 9:55 PM

Post by Diego Robledo on November 17, 2013

For Example #9, shouldn't the moles for Mg(OH)2 be 0.11 instead of 0.011? In addition, shouldn't 25.00mL Mg(OH)2, when converted to L, be 0.2500L instead of 0.02500L?

1 answer

Last reply by: Professor Franklin Ow
Thu Nov 7, 2013 5:17 PM

Post by HARRISON IGWE on September 8, 2013

calculate the volume of hydrogen produced when 6g of magnesium reacts with excess dilute hydrochloric acid at s.t.p

1 answer

Last reply by: Professor Franklin Ow
Thu Nov 7, 2013 5:19 PM

Post by HARRISON IGWE on September 8, 2013

calculate the mass of calcium oxide residue gotten when 82g of calcium carbonate is heated to a steady mass

Stoichiometry II

  • Stoichiometry uses coefficients from a balanced chemical equation as a conversion factor to relate any (2) reactants and/or products.
  • Mole to mole ratios are central to any stoichiometry problem.
  • The limiting reagent or reactant dictates how much product is expected to form (known as the theoretical yield).
  • Molarity expresses solution concentration, and can be used as a conversion factor.
  • Acid-base titrations are used to determine (or standardize) the concentration of an unknown solution.

Stoichiometry II

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Lesson Overview 0:10
  • Molarity 1:14
    • Solute and Solvent
    • Molarity
  • Molarity Cont'd 2:59
    • Example 1: How Many Grams of KBr are Needed to Make 350 mL of a 0.67 M KBr Solution?
    • Example 2: How Many Moles of KBr are in 350 mL of a 0.67 M KBr Solution?
    • Example 3: What Volume of a 0.67 M KBr Solution Contains 250 mg of KBr?
  • Dilutions 10:01
    • Dilution: M₁V₂=M₁V₂
    • Example 5: Explain How to Make 250 mL of a 0.67 M KBr Solution Starting From a 1.2M Stock Solution
  • Stoichiometry and Double-Displacement Precipitation Reactions 14:41
    • Example 6: How Many grams of PbCl₂ Can Form From 250 mL of 0.32 M NaCl?
  • Stoichiometry and Double-Displacement Precipitation Reactions 18:05
    • Example 7: How Many grams of PbCl₂ Can Form When 250 mL of 0.32 M NaCl and 150 mL of 0.45 Pb(NO₃)₂ Mix?
  • Stoichiometry and Neutralization Reactions 21:01
    • Example 8: How Many Grams of NaOh are Required to Neutralize 4.5 Grams of HCl?
  • Stoichiometry and Neutralization Reactions 23:03
    • Example 9: How Many mL of 0.45 M NaOH are Required to Neutralize 250 mL of 0.89 M HCl?
  • Stoichiometry and Acid-Base Standardization 25:28
    • Introduction to Titration & Standardization
    • Acid-Base Titration
    • The Analyte & Titrant
  • The Experimental Setup 26:49
    • The Experimental Setup
  • Stoichiometry and Acid-Base Standardization 28:38
    • Example 9: Determine the Concentration of the Analyte
  • Summary 32:46
  • Sample Problem 1: Stoichiometry & Neutralization 35:24
  • Sample Problem 2: Stoichiometry 37:50

Transcription: Stoichiometry II

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is going to be our second lecture on stoichiometry.0003

Let's go ahead and look at the lesson overview.0011

Last time we laid out a pretty good foundation for how to tackle stoichiometry problems.0014

We are just going to now build upon that foundation and apply stoichiometry to more specific cases.0022

These cases include the following; we are going first talk about solutions and their concentrations.0029

We are going to work with a lot of terminology today such as molarity and dilutions.0038

After discussing molarity and dilutions, the next specific application of stoichiometry is0046

going to be to double replacement precipitation reactions, acid-base neutralization, and finally0053

something we have not talked about yet which is called acid-base standardization via a titration.0062

After that we will go ahead and do our summary followed by our sample problems.0067

Moving on now, molarity.0077

We tend to think of a solution in chemistry as composed of two items--typically a solute and a solvent.0079

The solute is going to be present in the smaller quantity.0089

The solvent is going to be present in the major amount.0095

Of course for general chemistry lecture and for laboratory, the typical solvent of course is going to be water.0100

We always like to express a solution's concentration--basically how much solute we have relative to solvent.0111

There is a lot of different units we can use.0119

But the most common one we are going to use is called molarity.0121

The equation for molarity is defined as the following.0125

Molarity is equal to moles of solute for every liter of solution.0128

From this equation of course, we should always be very comfortable with solving for a single variable.0136

We can solve for liter of solution.0144

That is going to be basically moles of solute over the molarity.0148

Of course we can also solve for moles of solute.0157

Moles of solute is just going to be equal to molarity times liters of solution.0160

Once again you want to definitely have these equations committed to memory, especially for general chemistry.0172

Let's go ahead and jump right into a calculation.0181

How many grams of potassium bromide are needed to make 350 milliliters of a 0.67 molar KBr solution?0186

When we verbalize capital M which stands for molarity, this is verbalized as molar.0196

This is the language; this is the terminology that we use; 0.67 molar potassium bromide solution.0203

This is a first example.0210

We are just going to plug everything we know into the equation for molarity and solve for the unknown.0212

The molarity is given to us; that is 0.67 molarity.0217

That is equal to moles of solute divided by liters of solution.0223

We don't know what the moles of the solute is, the moles of KBr.0234

That is going to be what we are trying to find.0238

The liters of solution, you are actually told that in the problem.0241

It is here; it is 350 milliliters, which is 0.350 liters.0243

When we go ahead and solve for moles of potassium bromide, we get 0.2345 moles of KBr.0251

Of course you see that the question is not asking for moles, but the question is asking for grams.0262

Do you recall how to go from moles to grams?0268

What is the conversion factor called?--it is molar mass.0271

0.2345 moles of KBr times something over something.0276

That is going to give us the grams of KBr.0284

The molar mass is approximately 119 grams over 1 mole.0289

We are going to get an answer of approximately 28 grams of KBr.0294

We are really playing on the concept of the mole very heavily.0301

Once again what we learned last time in stoichiometry is going to serve as our foundation for the rest of this lecture.0306

So very important that we have those fundamentals down; let's go ahead.0312

What we just did was we used the molarity equation directly, moles over liters.0319

We just solve for the unknown.0323

But there is another way of using the molarity equation.0325

You see that the equation for molarity is basically a ratio of two units.0329

If you recall, I have said this many times before, that when you have a ratio of two units,0333

you can get a conversion factor that can be used in dimensional analysis.0338

Let's go ahead and use this approach now.0345

You see this is the exact same problem as before.0347

How many moles of KBr are in 350 milliliters of a 0.67 molar potassium bromide solution?--and how many grams is this?0350

Remember how we tackled dimensional analysis?--you always start with the unknown.0359

What is unknown?--what is the only thing that is given to us that is not part of a ratio yet?0366

350 milliliters is by itself; that looks pretty good.0372

0.67 molar, remember this is a ratio of two units.0374

This is really 0.67 moles for every 1 liter.0380

Or this is going to be 1 liter over 0.67 moles.0386

Remember this is how we use molarity as a conversion factor.0392

It is going to be moles over liters or liter over mole.0395

Let's start with what is given to us that is not part of a ratio.0401

That is going to be 350 milliliters.0404

You see that in the question for molarity, the volume is actually liters.0409

What do you think we have to do to 350 milliliters?0414

We have to get it to liters.0417

That is going to be 0.350 liters times something over something.0418

Let's go ahead and use our conversion factor.0425

We want liters downstairs; we want moles upstairs.0428

That is going to be 0.67 moles for every 1 liter.0432

That is going to give us our answer of 0.2345 moles of KBr.0437

From there you can go ahead and get the 28 grams of KBr on your own.0445

Again all this is is this is the exact same problem we just did about a minute ago.0450

But all we are doing is now we are using molarity as a conversion factor instead of just using the equation itself.0455

Either way, whatever is more comfortable for you, please use that way.0463

Let's go ahead and move on; how about volume?0468

What volume in milliliters of a 0.67 molar KBr solution contains 250 milligrams of KBr?0472

We see molarity here; we know that molarity is 0.67 moles over 1 liter.0480

We are not going to start with that because that is a ratio already.0488

The only other value is right here, 250 milligrams.0491

Remember what we have always done in stoichiometry.0496

We always wanted to get to moles; remember grams to moles.0499

That was so typical of being the first step in all of our problems, this lecture and last lecture.0506

Anytime you see mass in a stoichiometry problem, we are always going to get it to grams and then to moles.0515

0.250 grams of KBr, we are going to go ahead and get this to moles of KBr.0522

That is going to be times 1 mole divided by roughly 119 grams of KBr.0532

That gets us moles of KBr.0541

Now that we are in moles, we can use molarity as a conversion factor once again; the 0.67 moles over liters.0544

This time to get cancelled goes moles; liters goes upstairs; it is 0.67 in 1 liter.0552

That gives us our answer in liters after all is done and cancelled.0560

Now I want to get to milliliters of course; times 1 milliliter over 10-3 liters.0564

This is going to be an answer of 3.1 milliliters.0572

Once again any stoichiometry problem, always get to moles.0578

That always should be your first step.0588

You are going to hear me say that quite often.0593

Once again for stoichiometry, the first step is to get anything you can into moles.0596

The next application of stoichiometry, and this comes directly from molarity, is the practice of what we call the dilution.0604

When you order a chemical from a company, the chemical typically comes in the bulk form.0616

It is typically very concentrated; this is what we know as the stock solution.0623

When you receive the stock solution, you typically dilute it down to a desired concentration.0629

Being able to do a accurate and precise dilution is quite important in the laboratory setting.0636

To perform a dilution, the equation here is going to be used.0645

It is basically M1V1 equals to M2V2, where M1 is initial molarity and V1 is the initial volume.0649

M2 is final molarity; V2 is the final volume.0659

When you do a dilution, you typically just add water.0667

For example, sometimes parents when they give apple juice to their children, they typically add water to dilute it down.0671

Notice though that the only thing that they change is the amounts of solvent.0677

They do not change the amount of juice or sugar.0681

This amount of solute remains the same; amount of solute does not change in a dilution.0685

We can prove this mathematically.0698

Because when I take molarity which is moles over liters, and I multiply it by V1 which is liters, look what I get on M2V2.0701

M2 is moles over liters; and times liters; so moles is equal to moles.0710

Once again in a dilution, the moles of solute before equals the moles of solute after.0716

We are now going to get into a typical dilution problem.0726

Explain how to make 250 milliliters of a 0.67 molar potassium bromide solution starting from a 1.2 molar stock solution.0731

Let's go ahead and write out the equation again; M1V1 equals to M2V2.0740

Let's see what we can plug in now; M2V2 is already given to us.0748

M2 is going to be the desired concentration of 0.67 molar.0752

V2 is the desired volume; that is going to be 250 milliliters or 0.250 liters.0758

The M1 is given to us also; it is the 1.2 molar stock solution.0769

This V1 therefore is the unknown; it is pretty simple to solve for this algebraically.0776

After we are going to do this, we are going to then discuss the physical significance of this because that is another issue.0785

V1 is going to be 139.6 milliliters; what is the physical significance of this volume?0790

This is the amount of the stock solution you are actually going to take; amount of stock solution to dilute.0801

But how much water do you add then to dilute it?0815

If we want 250 milliliters as the final volume, and we are going to0819

use 139.6 milliliters of the stock, all I have to do is subtract.0827

Vfinal minus Vinitial is going to give me the volume of water to add to perform the dilution.0837

That is going to be equal to 110.4 mL of water.0850

If you were to do this physically in the laboratory, you will literally take 139.6 milliliters of stock solution.0858

To it, you would add 110.4 milliliters of water.0868

That is going to give you 250 milliliters of the 0.67 molar potassium bromide solution.0871

Once again this is how you do a standard typical dilution problem.0878

Let's move on to the next application of what we learned last time in stoichiometry.0885

The good news about this is it is going to be all familiar.0891

This is stoichiometry and double displacement precipitation reactions.0893

Before we tackled the following questions.0900

How many grams of lead chloride can form from 50 grams of NaCl?0903

We know how to do this.0908

We went from grams of A to moles of A; then to moles of B; then on to grams of B.0911

Remember this was what we called a mass to mass conversion; mass to mass conversion.0921

Now we are going to change it up a little.0932

All we are doing is we are asking the same problem where now we are going to incorporate molarity.0934

Now how many grams of lead(II) chloride can form from 250 milliliters of 0.32 molar NaCl?0939

Remember what is always the first step in a stoichiometry problem is to get to moles; get to moles.0949

We have 250 milliliters; we have 0.23 molar; we have a volume and a molarity.0962

We are going to take 0.250 liters; we are going to multiply that by something over something.0969

That is going to give us moles of sodium chloride.0975

Liters goes downstairs; moles goes upstairs; that is going to be 0.32 over 1.0980

That is going to give us moles of sodium chloride.0988

Now what we want to do again is grams of A to moles of A, to moles of B, to grams of B.0993

We are still going to follow that flow chart.1002

This is really our moles of A.1005

Now we are going to get to moles of B.1009

We are going to take x moles of NaCl.1011

Remember how we get to the mole to mole ratio?1017

It comes from the balanced chemical equation; 2 sodium chlorides to 1 lead(II) chloride.1019

That is going to be times 1 mole of lead(II) chloride for every 2 moles of sodium chloride.1026

That is moles of B; now from moles of B, we go on to grams of B.1037

Multiply by the molar mass of lead(II) chloride which is 278.1 grams of PbCl2 for every 1 mole of PbCl2.1041

When all is said and done, we get 11.1 grams of PbCl2.1051

This is a very typical problem; don't forget what we did here was the following.1058

Basically volume times molarity gets you the moles.1064

I cannot underscore this enough of how often you are going to use this when you do a stoichiometry problem.1069

Remember volume times molarity is equal to moles.1077

It is going to be typically your first step.1080

Let's do another problem.1088

How many grams of lead(II) chloride can form when 250 milliliters of1090

0.23 molar sodium chloride and 150 milliliters of 0.45 molar lead(II) nitrate mix?1096

You notice that you are given amounts of both reactants instead of just one.1104

Remember what type of problem this is anytime you are given amounts of both reactants?1111

That is right; it is a limiting reactant problem.1116

The limiting reactant problem that we learned in the previous lecture can easily still apply to what we learn today.1119

Let's go ahead and do it.1126

Remember we are going to use what was called the smaller quantity method to solve limiting reactant problems.1128

Let's go ahead and do this.1141

Remember we are going to determine the amounts of product that can form from each of the reactant's amounts.1142

0.250 liters, this is going to be times 0.32 moles over liter of sodium chloride.1149

We want to get to lead chloride; remember it is a 1:2 ratio.1161

We just did this; that is going to give us 0.040 moles of lead(II) chloride.1164

I am going to repeat the process now for the other reactant amount.1176

0.150 liters times 0.45 moles of lead(II) nitrate over 1 liter of the lead(II) nitrate.1180

When you look back at the balanced chemical equation, the mole to mole ratio between lead(II) chloride and lead(II) nitrate is 1:1.1197

Times 1 mole of lead(II) chloride divided by 1 mole of the lead(II) nitrate.1205

When that is all calculated, you get 0.68 moles of lead(II) chloride.1214

In this specific example, the theoretical yield should be 0.040 moles of lead(II) chloride.1221

That translates to 11.1 grams of lead(II) chloride.1228

Therefore the limiting reactant in this specific example is sodium chloride.1232

Again we are using previously learned material in today's lecture; that is all.1244

It is just a slight twist on things, but you are totally capable of doing this.1249

Again it is just using old material.1252

That is the nice thing about this chapter, about this lecture.1254

We are really building upon what we have learned just recently.1257

The next specific application of stoichiometry is going to be to again another aqueous reaction that we have already talked.1265

That was the acid-base neutralization reaction.1274

If everybody recalls, in a neutralization reaction, it is always between an acid and a base.1277

The two products are always going to be water and a salt.1283

Again a neutralization reaction between a Bronsted-Lowry acid and a base is just another double replacement reaction.1288

This equation is balanced; so we are good to go and ready to use it.1297

How many grams of sodium hydroxide are required to neutralize 4.5 grams of hydrochloric acid?1301

Once again grams of A to moles A, goes on to moles of B, goes on to grams of B.1307

This is a typical mass to mass conversion again.1318

Don't get thrown off by the word neutralize.1321

It is just another term there that means it is going to react with each other.1324

All again this is some mass to mass conversion.1328

4.5 grams of hydrochloric acid times 1 mole divided by 36.458 grams of hydrochloric acid.1332

That is its molar mass roughly; that gives us the moles of A.1343

Moles of A to moles of B, that is going to be a 1:1 ratio.1349

That is 1 mole of sodium hydroxide over 1 mole of the HCl.1352

Finally now we are in moles of B.1359

Let's go ahead and move on to grams of B.1362

That is going to be times 39.998 grams of sodium hydroxide over 1 mole.1364

4.9 grams of sodium hydroxide are required for this reaction to proceed, are required to neutralize the indicated amount of hydrochloric acid.1372

We are going to tackle the same problem.1386

But now we are going to apply molarity to it.1388

Because this is aqueous after all; so we can use molarity anytime.1391

How many milliliters of 0.45 molar sodium hydroxide are required to neutralize so much of HCl?1395

Remember what is always the first step?1404

It is to get anything you can into moles.1406

For sodium hydroxide, all I am given is the molarity.1410

I cannot get that into moles because that says moles over liters.1412

But you see that combination again for hydrochloric acid, the volume and molarity.1416

That is always a one-two combo.1421

Anytime you see molarity and volume, multiply the two together.1424

You are going to get moles; don't forget that.1428

You are going to hear me say that again.1432

Don't forget that; this is so repetitive.1433

You can totally feel confident in doing this because it is the same tools over and over again.1436

0.250 liters of the hydrochloric acid times 0.89 moles of hydrochloric acid over 1 liter of hydrochloric acid.1443

That is moles of A; now from moles of A on to moles of B.1456

That is a 1:1 ratio from the balanced chemical equation.1460

Times 1 mole of sodium hydroxide divided by 1 mole of the hydrochloric acid.1464

That gets us... you know what?--we have to do one more step.1471

We still have to use the molarity there; my apologies.1484

Moles of sodium hydroxide goes downstairs; liters goes upstairs.1490

That is going to be 0.45; and 1 on top.1493

When all is said and done, we get 0.494 liters of sodium hydroxide required.1497

The question is asking for milliliters.1503

That is going to be 494 milliliters of sodium hydroxide required.1505

This is a nice cumulative problem because it really utilizes the concept of molarity as a conversion factor. not once but twice.1511

Again this is stoichiometry and neutralization reactions.1526

The next specific application of stoichiometry pertains also to acid-base chemistry.1531

This is what we call the concept of standardization.1538

Standardization is basically determining the concentration of an unknown.1542

To do this, we use the process known as titration.1548

Titration is a general term where a solution of known concentration is used to determine the concentration of an unknown solution.1552

In other words, we are going to standardize the unknown.1560

We are going to find out its exact concentration.1562

There are many different types of titration.1566

A very common and specific type is also acid-base titration.1568

In acid-base titration, a basic or acidic solution of known concentration is used to determine the concentration of an acidic or basic solution.1572

The analyte is what we are trying to find; this is the unknown.1585

The titrant is what we are going to be using to standardize the unknown.1594

This is the known concentration; this is the solution of known concentration.1598

Let's get into the experimental setup; a typical titration has the following.1610

A burette is basically a very precise piece of glassware used to deliver small aliquots of liquid.1615

Burettes are usually pretty precise up to 0.01 milliliters.1625

In the burette typically goes the titrant; inside the flask then goes two things.1631

It goes the analyte and also what we call an indicator.1645

What an indicator's job is is the following.1651

It is going to provide a visual sign of when the titration is complete.1653

We are going to talk about titrations, more complicated ones, down the line.1669

But for now an indicator's job again is to tell us when the titration is complete.1675

Remember this is acid-base neutralization after all.1682

This is going to tell us when the neutralization has occurred; complete neutralization has occurred.1685

The visual sign is the indicator is going to change color.1700

Again you are going to do this so much in general chemistry lab, I promise you.1709

It is one of the most traditional types of experiments we teach undergraduates.1712

Let's go ahead and jump into a typical acid-base standardization problem.1719

We have to get familiar with the terminology.1724

32.10 milliliters of 0.67 molar HCl was required to reach the equivalence point with 25 milliliters of magnesium hydroxide.1727

Determine the concentration of the analyte.1736

First let's get some terminology out of the way.1739

The equivalence point is the point in the titration where complete neutralization has occurred.1741

The point in the titration where complete neutralization has occurred.1751

This is very important that you pay attention to these key words.1764

Because that tells you pretty much how much has reacted, when the reaction has complete.1768

Step one, you always hear me... that you always want to get the moles as step one.1777

That is usually true but what is even more important before that is that we have a balanced chemical equation.1790

Anytime you are doing a stoichiometry problem, make sure you know the balanced chemical equation or get it.1797

Let's go ahead; this is HCl aqueous plus MgOH2 aqueous.1805

That is going to go ahead and form H2O liquid and MgCl2 aqueous.1815

We have to go ahead and balance it.1823

This is going to be 2 hydrochloric acids and 2 waters; why is that important?1825

Because you just know that we are going to be using a mole to mole ratio from the balanced chemical equation.1831

So it is imperative that you know the coefficients correctly.1836

Step two, get anything you can into moles; get all amounts possible into moles.1841

We have the volume of magnesium hydroxide; we can't do anything with that yet.1857

The only thing here is hydrochloric acid.1861

You see that we are given both the volume and molarity; don't forget.1865

You see that combination again; volume times molarity is going to get you moles.1869

Let's go ahead and do this; 0.3210 liters times 0.67 moles of HCl over 1 liter.1873

That gets us to moles of A.1886

Once we get in moles of A, let's go ahead and get to moles of B.1889

It is a 2:1 ratio; 1 magnesium hydroxide on top divided by 2 moles of hydrochloric acid on the bottom.1893

That is going to give us 0.011 moles of MgOH2.1905

The question is asking for the concentration of the analyte, the concentration of magnesium hydroxide.1913

The only other value in the problem we haven't used yet is the 25 milliliters.1920

Remember that we want concentration which is molarity; molarity is moles over liters.1926

We know the liters; it is given to us in the original problem.1932

There is 25 milliliters; that is going to be 0.02500 liters.1938

The moles of magnesium hydroxide we just solved for.1944

When all is said and done, we are going to get an answer of 0.43 molar MgOH2.1949

This is how you determine the concentration of unknown.1959

This is how we standardize it using the concentration of a known substance.1962

Let me go ahead and summarize today's lecture then.1969

The first take-home message is that when trying to define the concentration of a solution, the most common unit used is molarity.1972

Again molarity is moles of solute over liter of solution.1982

Molarity can be used as a conversion factor in dimensional analysis problems including1986

dilutions problems, double replacement precipitation reactions, acid-base neutralization, and acid-base standardization via titration.1991

Finally let's go over the basic steps and formats for doing general stoichiometry problems.2002

Step one is to make sure you come up with the balanced chemical equation if not provided for you already; balanced chemical equation.2009

Step two, step two is to get all amounts you can into moles; get all amounts possible into moles.2025

Remember there are some common patterns that we have seen already.2038

Usually we can go from grams to moles.2042

That is going to be via molar mass.2046

But another thing we have seen is using molarity.2050

That was volume times molarity is going to get you moles.2052

Again that is using molarity as a conversion factor.2057

The first one here we use molar mass as a conversion factor.2068

That is going to get us into moles of A.2077

Then once you are in moles of A, you can go on to moles of B.2084

From moles of B, you can go on then to solve the problem.2090

Sometimes the question asks us for grams of B.2094

Sometimes the question asks us for volume of B.2098

Again if we are going to do grams of B, once again that is going to be molar mass as a conversion factor.2104

But if we are going to go to volume of B, we are going to be using simply molarity again as a conversion factor.2111

I hope that is a nice little summary of how to tackle general stoichiometry problems.2119

Let's now work on some more sample problems.2124

Sample problem number one, how many milliliters of 0.45 molar sodium hydroxide are required to neutralize 250 milliliters of 0.89 molar sulfuric acid?2129

Step one, we don't have any balanced chemical equations.2141

Let's go ahead and come up with it on our own; sulfuric acid reacting with sodium hydroxide.2145

That is going to go ahead and give us water and our salt, Na2SO4 aqueous.2156

Let's go ahead and balance it.2165

We are going to be needing two of these and of course two of those.2168

Step one is done.2174

Step two is to get any quantities you can into moles; get into moles.2176

Remember what are the possible combinations?2185

Usually we go from grams to moles via molar mass.2189

But you don't see any grams in this problem.2192

The other alternative is to do volume times molarity getting you into moles.2194

Here we have the volume and molarity for sulfuric acid.2198

That is going to be 0.250 liters times 0.89 moles of sulfuric acid over 1 liter.2202

That gets us into moles of A; now we want to go to moles of B.2213

The mole to mole ratio is 1 sulfuric acid to 2 sodium hydroxides.2225

Times 2 moles of sodium hydroxide for every 1 mole of sulfuric acid.2229

Finally the question is asking for volume; we want to get now into milliliters.2237

To do that, we use molarity; times 1 liter over 0.45 moles.2242

That is going to give us a answer of 0.99 liters of the sodium hydroxide which is2250

going to be 990 milliliters required for neutralization to occur of the specified amount of sulfuric acid.2258

That is sample problem number one; let's now move on to sample problem number two.2269

How many grams... right away I see grams; that is mass.2273

So I know for sure that I am going to be using probably molar mass as a conversion factor.2279

Mass via molar mass as conversion factor.2285

How many grams of barium sulfate can form when so much sodium sulfate reacts with so much barium nitrate?2291

Let's go ahead; we see that we have both amounts of two reactants.2299

So you know that this is a limiting reactant problem.2306

Let's go ahead and go through it.2316

We are going to be using the smaller quantity method.2317

Step one, get the balanced chemical equation if it hasn't been provided to you already.2327

Here we have to come up with it; balance chemical equation.2333

It is going to be sodium sulfate aqueous reacting with barium nitrate aqueous.2342

That is going to go on and form barium sulfate solid and sodium nitrate aqueous.2353

Let's go ahead and balance this guy.2363

I am going to be needing two of those for our balanced chemical equation.2365

Alright, step one is complete.2370

Step two... you guys know the drill.2373

Step two is to get all quantities you can into moles; get into moles.2376

Look here; for both of the reactants, for sodium sulfate we have both the volume and the molarity.2382

For barium nitrate, we have the volume and the molarity.2390

So volume times molarity gets us into moles.2393

We are going to do this for each reactant.2398

We are going to see which one gives us the smaller theoretical yield for the product.2400

Let's work with sodium sulfate first.2407

0.250 liters times 0.32 moles of sodium sulfate over 1 liter.2409

That gets us into moles of A; now we go on to moles of B.2419

The mole to mole ratio between sodium sulfate and barium sulfate is 1:1.2423

1 mole of barium sulfate over 1 mole of sodium sulfate.2429

That is going to give us 0.080 moles of barium sulfate.2437

We are going to continue on with the smaller amount method for solving the limiting reactant problem.2444

We are going to repeat the process for the other reactant amount.2449

0.125 liters times 0.87 moles of barium nitrate divided by 1 liter.2452

The mole to mole ratio between barium nitrate and barium sulfate is also 1:1.2463

That is 1 mole, barium sulfate over 1 mole of barium nitrate.2469

That is going to give us 0.11 moles of barium sulfate.2478

Using our smaller amount method, the theoretical yield is going to be 0.080 moles of product.2488

That is going to be 18.7 grams of barium sulfate formed, which means sodium sulfate is the limiting.2496

Again this is a limiting reactant problem using molarity as a conversion factor.2511

I hope you leave today's lesson feeling pretty confident and comfortable doing a stoichiometry problem.2515

It seems like a lot of calculations.2523

But if you follow the basic steps, you will get it every time.2524

Remember step one is balance chemical equation if not given to you already.2528

Step two is going to be basically getting all the quantities you can into moles.2533

If you need to go from grams to moles, it is always molar mass.2538

If you need to go from volume to moles, it is always going to be molarity for these two lectures.2543

Thank you for using Educator.com again; I will see you again next time.2554

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