Stoichiometry and Molarity Review

Key Announcement

Stoichiometry Concepts

• Basic Stoichiometry: Conversion between moles of substances using coefficients from balanced equations.

  • Example 1: Converting moles of water to moles of oxygen.

    • Given: 2.33 moles of $H_2O$

    • Coefficients: $H2O = 4$, $O2 = 6$.

    • Calculation:

    • Moles of $O2 = (2.33 ext{ moles of } H2O) * ( rac{6 ext{ moles of } O2}{4 ext{ moles of } H2O}) = 3.495 ext{ moles of } O_2$.

• Moles to Mass Conversion: Multiply moles by molar mass.

  • Example 2: Convert moles of oxygen to grams.

    • 1.55 moles of $O_2$, Molar mass = 32 g/mol.

    • Calculation: 1.55 moles * $32 g/mol$ = 49.6 g $O_2$.

• Converting Grams to Moles: Useful for determining moles from a mass of a compound.

  • Example 3: Given 58.6 g of $A NO_3$, convert to moles of oxygen.

    • Use molar mass of $A NO_3$ (101.11 g/mol).

    • Calculation:

    • Moles = rac{58.6 g}{101.11 g/mol} = 0.579 moles.

    • Use coefficients $O2 = 5$, $A NO3 = 3$.

    • Calculation: Final moles of $O_2 = (0.579 moles) * rac{5}{3} = 0.965 moles$.

Limiting Reactant Problems

• Identifying Limiting Reactant: Begin with two reactants and find the limiting one by calculating the products for each reactant.

  • Example: Given 4 moles each of $Al2S3$ and $H_2S$, find which produces the lesser amount of product (hydrogen sulfate).

    • Calculation: $Al2S3$ to $H2S$ yields 12 moles; $H2O$ to $H_2S$ yields 2 moles.

    • Choose the smaller number (2 moles) as limiting reactant.

Percent Yield Calculations

• Calculating Percent Yield: Percent yield = (Actual Yield / Theoretical Yield) x 100.

  • Example: Given actual yield of $CO_2$, calculate theoretical yield using limiting reactant.

    • If theoretical yield calculated as 2.56 g,

    • Actual Yield = 1.9 g,

    • Percent Yield = rac{1.9}{2.56} * 100 = 74.22%.

Molarity Problems

• Understanding Molarity: Molarity (M) = Moles of solute/Volume of solution in liters.

  • Example: Given molarity and volume, solve for moles.

  • Example 1: Molarity = 0.733 M, Volume = 548 mL, convert mL to L (0.548 L).

  • Calculation: Moles = Molarity * Volume = $0.733 imes 0.548$ = 0.402 moles.

Conductivity and Solubility

• Conductivity: Determined by concentration of ions in solution.

  • Example: Different solutions compared based on number of ions.

    • Compound A: 2 Cl and 3 SO$_4 = 5$ ions total.

    • Compound B: 2 Cl and 1 CO$_3 = 3$ ions total.

    • Answer: Compound A has higher conductivity.

Oxidation-Reduction Reactions (Redox)

• Oxidation and Reduction: Remember: oxidation is the loss of electrons, reduction is the gain of electrons.

  • Example: Determine oxidation states and changes.

    • Chromium oxidation state: Reduction from +7 to +2 (inferred from equation).

    • Oxidation Labeling: Identify changes in oxidation state to find which elements are oxidized/reduced.

Review Problems

• Practice problems from transcribed issues regarding stoichiometry, molarity, conductivity, and redox reactions.