Unit 6

Unit 6 Study Guide: Chemical Reaction Stoichiometry

This study guide covers the fundamental principles of chemical reaction stoichiometry, including defining key terms, understanding quantitative relationships in reactions, identifying limiting reactants, calculating yields, and applying these concepts to solutions.

I. Key Definitions

  • Reactants: Substances undergoing chemical change.

  • Products: Substances formed from a chemical reaction.

  • Chemical Reaction: Transformation of one set of chemical substances to another.

  • Chemical Equation: Symbolic representation of a chemical reaction (A + B \rightarrow C + D).

  • Stoichiometry: Study of numerical relationships among chemical quantities in balanced reactions.

  • Limiting Reactant (Reagent): Reactant consumed first, limiting product formation.

  • Theoretical Yield: Maximum product achievable from given reactants.

  • Solution: Homogeneous mixture of solute and solvent.

  • Solute: Component being dissolved.

  • Solvent: Component doing the dissolving.

  • Solution Concentration: Measure of solute amount relative to solvent.

  • Molarity (M): Moles of solute per liter of solution (M = \frac{\text{moles of solute}}{\text{liters of solution}})

II. Core Concepts & Principles

  1. Law of Conservation of Mass:

    • Atoms rearrange, but the number of each type of atom remains constant from reactants to products.

  2. Balancing Chemical Equations:

    • Atoms combine in whole-number ratios.

    • Coefficients must be whole numbers and reduced to the lowest possible values.

    • Physical states: (s) solid, (l) liquid, (g) gas, (aq) aqueous.

    • Example: N2(g) + 3H2(g) \rightarrow 2 NH_3(g)

      • This means 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH_3.

III. Stoichiometric Calculations
A. General Steps for Stoichiometric Calculation

  1. Write and balance the chemical equation.

  2. Determine molar masses of substances involved.

  3. Use coefficients from the balanced equation for mole conversions (stoichiometric ratio).

  4. Convert moles of the desired substance to grams (or vice versa) using molar mass.

B. Example: Mole-to-Mole Conversion

  • Problem: How many moles of CO2 are formed if 22.0 moles of C8H_{18} are combusted?

  • Balanced Equation: 2 C8H{18}(l) + 25 O2(g) \rightarrow 16 CO2(g) + 18 H_2O(g)

  • Calculation: 22 \text{ moles } C8H{18} \times \frac{16 \text{ moles } CO2}{2 \text{ moles } C8H{18}} = 176 \text{ moles } CO2

C. Example: Mass-to-Mass Conversion

  • Problem: Determine mass of CO2 produced when 3.6 \times 10^{15} grams of C8H_{18} are burned.

  • Molar Masses: MM(C8H{18}) = 114.2 \text{ g/mol}, MM(CO_2) = 44.01 \text{ g/mol}

  • Strategy: Grams C8H{18} \rightarrow Moles C8H{18} \rightarrow Moles CO2 \rightarrow Grams CO2

IV. Limiting Reactants and Yield
A. Concepts

  • Limiting Reagent: The reactant completely consumed first, dictating the maximum amount of product.

  • Excess Reactants: Reactants present in amounts greater than needed.

  • Theoretical Yield: The maximum amount of product that could be formed from the limiting reactant.

  • Actual Yield: The experimentally obtained amount of product.

B. Percent Yield

  • Formula: \text{Experimental Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%

  • Units for actual and theoretical yield must match (e.g., grams).

  • Actual yields are often less than theoretical due to incomplete reactions, side reactions, or losses.

C. Example: Percent Yield Calculation

  • Problem: 50.0 \text{ g} of Cu_2S reacts with 10.0 \text{ g} of C yielding 33.5 \text{ g} of Cu.

  • Balanced Reaction: 2 Cu2S + C \rightarrow CS2 + 4 Cu

  • Theoretical Yield Calculation (assuming Cu_2S is limiting):

    • 50.0 \text{ g } Cu2S \times \frac{1 \text{ mol } Cu2S}{159.2 \text{ g}} \times \frac{4 \text{ mol } Cu}{2 \text{ mol } Cu_2S} \times \frac{63.55 \text{ g } Cu}{1 \text{ mol } Cu} = 39.9 \text{ g } Cu

  • Percent Yield: \frac{33.5 \text{ g } Cu}{39.9 \text{ g } Cu} \times 100\% = 84.0\% (Note: The original calculation in the provided text had a minor rounding discrepancy in 84.57%, corrected to 84.0% for consistency with the example numbers provided)

V. Solution Stoichiometry
A. Molarity and Solution Preparation

  • Molarity (M): Moles of solute per liter of solution.

  • Preparing a solution: Weigh solute, dissolve in solvent, then add more solvent to reach the desired total volume.

  • Example: To prepare 100 mL of 0.050 M Ca(ClO)_2

    • 0.1 \text{ L} \times 0.050 \text{ mol/L} \times 142.9 \text{ g/mol} = 0.715 \text{ g } Ca(ClO)_2

B. Ionic Separation in Solution

  • When ionic compounds dissolve, they dissociate into ions.

  • Example: Na2S2O3 \rightarrow 2 Na^+ + S2O_3^{2-}

    • If [Na2S2O3] = 0.127 \text{ M}, then [Na^+] = 2 \times 0.127 \text{ M} = 0.254 \text{ M} and [S2O_3^{2-}] = 0.127 \text{ M}.

C. Dilution from Stock Solution

  • Adding solvent to a concentrated stock solution.

  • Moles of solute remain constant: M1V1 = M2V2

    • Where M is molarity and V is volume.

VI. Key Takeaways for Review

  • Be able to balance chemical equations accurately.

  • Identify limiting reagents and excess reactants in a given reaction.

  • Calculate theoretical yield and percent yield.

  • Perform solution stoichiometry calculations, especially using molarity and dilution principles.