2/26/25 CCP Chemistry Lecture (Unit 2.3, Chap 5)

Agenda for Today's Class

  • Review molarity examples and solve problems.

  • Discuss aqueous reactions including acid-base and gas-forming reactions.

  • Highlight crossovers from Unit 1 material that are relevant for Unit 2.

The Importance of Prior Material

  • Material from Unit 1 is essential for understanding Unit 2, especially in this chapter and Chapter 4.

  • Reinforce understanding of unit conversion and stoichiometry from prior chapters:

    • Encourage students to revisit Chapter 3 for clarification.

Molarity and Stoichiometry Example

Reaction Overview

  • Reaction: Sulfuric acid (H₂SO₄) reacts with sodium hydroxide (NaOH) to produce sodium sulfate (Na₂SO₄) and water (H₂O).

  • Type: Neutralization reaction (acid + base = salt + water).

Calculation Steps

  • Given:

    • 100 mL of 1 M H₂SO₄

    • How much NaOH is required? (25 mL)

  • Objective: Calculate the molarity of the sodium hydroxide solution using titration principles.

    • Molarity = moles of solute / liters of solution.

Determining Moles of NaOH

  • Convert volume of NaOH from mL to liters (25 mL = 0.025 L).

  • Use given molarity of sulfuric acid to find moles:

    • 1 M sulfuric acid means 1 mole/L.

    • Calculate moles of H₂SO₄:

      • Moles = 1 M x 0.1 L = 0.1 moles.

  • Stoichiometry:

    • Reaction shows a 2:1 mole ratio (2 moles of NaOH for every 1 mole of H₂SO₄).

    • Moles of NaOH = 2 x moles of H₂SO₄ = 0.2 moles.

  • Finally, calculate molarity of NaOH:

    • Molarity (NaOH) = moles / liters = 0.2 moles / 0.025 L = 8 M.

Second Example: Lithium Carbonate and Hydrochloric Acid Reaction

Reaction Overview

  • Reaction: Lithium carbonate (Li₂CO₃) reacts with hydrochloric acid (HCl) to produce lithium chloride (LiCl), CO₂ and water (H₂O).

Calculation Steps

  • Given:

    • 50 g of lithium carbonate

    • 250 mL of 0.35 M hydrochloric acid.

  • Limiting Reagent Determination:

  1. Convert grams of Li₂CO₃ to moles using molar mass (73.89 g/mol).

  2. Use molarity of HCl to find moles:

    • 0.35 M means 0.35 moles/L. Therefore, Moles of HCl = 0.35 M x 0.250 L = 0.0875 moles.

  3. Determine the limiting reagent by comparing moles needed for complete reaction.

    • Based on stoichiometry (1 Li₂CO₃ : 2 HCl), calculate that 0.0875 moles HCl require:

      • Need for 0.0875/2 = 0.04375 moles of Li₂CO₃, and enough is present (50 g).

  • Therefore, HCl is limiting reagent.

  • Final Calculation for Products:

    • Calculate grams of lithium chloride formed:

      • 0.0875 moles of HCl yield equimolar LiCl (0.0875 moles).

      • Molar mass of LiCl is ~42.39 g/mol. Grams of LiCl = 0.0875 mol x 42.39 g/mol = 3.71 grams.

Key Takeaways on Aqueous Reactions

  • Emphasize the crossover method in predicting products from reactants.

  • Discuss solubility and phase tendencies of compounds in reactions to assist in identifying products.

  • Illustrate balancing chemical equations through examples.

Study Recommendations

  • Encourage reviewing previous chapters' materials and solved problems.

  • Practice with stoichiometry principles and conversions (grams to moles, liters to moles).

  • Highlight the need to distinguish limiting and excess reagents in reactions.

  • Net ionic equations will be covered next, requiring understanding of dissociation in aqueous solutions.