Chem 1
Chapter 1: Introduction to Limiting Reactants and Stoichiometry
Balanced Reaction Example
Reaction: Zinc + H₂SO₄ → ZnSO₄ + H₂
Coefficients are all 1 when not explicitly stated.
Regular Stoichiometry vs. Limiting Reactants
Regular Stoichiometry: One given value (e.g., 10g of Zinc) to find another single value (e.g., grams of H₂).
Limiting Reactant Problem: Multiple given values, need to identify which reactant runs out first (the limiting reactant).
Theoretical Yield
Maximum amount that could be produced from a reaction, if everything goes perfectly.
Steps for Limiting Reactant Problem
Set up calculations using both reactants.
Convert given amounts to moles.
Use stoichiometric ratios from the balanced equation to find moles of desired product.
Convert moles back to desired units (grams, etc.).
Calculating Example
Using Zinc (Zn)
Convert 10g Zn to moles:
Grams to moles: 10 grams Zn * (1 mole Zn / 65.4 grams) = 0.153 moles Zn.
Convert moles Zn to moles H₂:
0.153 moles Zn * (1 mole H₂ / 1 mole Zn) = 0.153 moles H₂.
Convert moles H₂ to grams:
0.153 moles H₂ * (2 grams H₂ / 1 mole H₂) = 0.306 grams H₂.
Using H₂SO₄
Convert 10g H₂SO₄ to moles:
10 grams SO₄ * (1 mole H₂SO₄ / 98 grams) = 0.102 moles H₂SO₄.
Convert moles H₂SO₄ to moles H₂:
0.102 moles H₂SO₄ * (1 mole H₂ / 1 mole H₂SO₄) = 0.102 moles H₂.
Convert moles H₂ to grams:
0.102 moles H₂ * (2 grams H₂ / 1 mole H₂) = 0.204 grams H₂.
Identifying Reactants
Limiting Reactant: The one that produces a lesser amount of product (0.204 grams H₂ from H₂SO₄).
Excess Reactant: The reactant that remains (Zinc).
Chapter 2: Percent Yield
Definition of Percent Yield
Evaluates the success of a reaction:
Percent Yield = (Actual Yield / Theoretical Yield) * 100%
Example Calculation
Theoretical Yield: 0.204 grams H₂; Actual Yield: 0.150 grams H₂.
Percent Yield Calculation: (0.150 g / 0.204 g) * 100% = 73.5%.
Contextual Insights
90-95% is a good yield in general chemistry; lower yields might indicate issues.
Chapter 3: Aqueous Solutions and Ionic Compounds
Definition of Aqueous Solutions
Aqueous (AQ): Substance dissolved in water.
Components: Solute (part being dissolved) & Solvent (part doing the dissolving).
Concentrations
Dilute Solutions: Less solute.
Concentrated Solutions: More solute.
Ionic Compounds in Water
Soluble: Break into ions and conduct electricity.
Insoluble: Form precipitates (represented as PPT).
Net Ionic Equations
Molecular Equation: Whole equation as written.
Ionic Equation: Breaks down soluble compounds into ions.
Net Ionic Equation: Shows only what participates in the reaction, eliminating spectator ions.
Chapter 4: Solubility Rules and Precipitation Reactions
Solubility Chart
Group 1A elements and nitrates are always soluble.
Specific ions (Ag⁺, Pb²⁺, and Hg²⁺) create insoluble compounds with certain anions.
Double Replacement Reactions
Identify products by mixing cations and anions from reactions.
Confirm products' solubility using solubility rules to predict precipitate formation.
Chapter 5: Molarity and Dilution
Definition of Molarity
Molarity (M) = Moles of solute / Liters of solution.
Calculating Dilutions
Dilution Equation: M₁V₁ = M₂V₂.
Maintain paired values (molarity and volume).
Initial values (M₁ and V₁) vs. final values (M₂ and V₂).
Chapter 6: Problem Solving Using Molarity
Examples on Conversion
Converting grams to moles using molecular weight.
Understanding the importance of keeping units consistent during calculations.
Chapter 7: Application of Stoichiometry through Solutions
Problem Example
Preparing from given molarity to find mass of product generated.
Using molarity to calculate how many grams of a substance could be produced based on molarity and volume.
Chapter 8: Conclusion and Key Concepts
Understanding Solubility and Reactions
Recognize how to identify solubility and conduct net ionic reactions.
Comprehend definition and application of percent yields in evaluating reaction efficiency.
Revisit stoichiometry principles in context with molarity and dilutions.