Calculating Reaction Yields

Background: Theory vs. Reality

Theoretical Yield: The maximum amount of product that can be produced based on stoichiometry.

Actual Yield: The amount of product physically obtained in the lab.

Percent Yield:

Percent Yield = Actual Yield Theoretical Yield × 100%

Problem 1

Step 1 (Part A): Determining Required Reactants

Step 1: Convert 45.0 g CaCl₂ to moles

45.0 g CaCl₂ × 1 mol CaCl₂ 110.98 g CaCl₂ = 0.405 mol CaCl₂

Step 2: Determine moles of Ca required

0.405 mol CaCl₂ × 1 mol Ca 1 mol CaCl₂ = 0.405 mol Ca

Convert moles Ca to grams

0.405 mol Ca × 40.08 g Ca 1 mol Ca = 16.2 g Ca

Step 3: Determine moles of Cl₂ required

0.405 mol CaCl₂ × 1 mol Cl₂ 1 mol CaCl₂ = 0.405 mol Cl₂

Convert moles Cl₂ to grams

0.405 mol Cl₂ × 70.90 g Cl₂ 1 mol Cl₂ = 28.7 g Cl₂

Step 2 (Part B): Calculating Actual Yield

Percent Yield = 39.6 g 45.0 g × 100 = 88.0%

Problem 2

Balanced Chemical Equation

The chemist wants to produce 100.0 g of Al₂O₃. Determine the moles and grams of Al and O₂ needed.

Step 1 (Part A): Determining Required Reactants

Step 1: Convert grams Al₂O₃ to moles

100.0 g Al₂O₃ × 1 mol Al₂O₃101.96 g Al₂O₃ ≈ 0.980 mol Al₂O₃

Step 2: Determine moles of Al required

0.980 mol Al₂O₃ × 4 mol Al2 mol Al₂O₃ = 1.960 mol Al

Convert moles Al to grams

1.960 mol Al × 26.98 g Al1 mol Al ≈ 52.9 g Al

Step 3: Determine moles of O₂ required

0.980 mol Al₂O₃ × 3 mol O₂2 mol Al₂O₃ = 1.470 mol O₂

Convert moles O₂ to grams

1.470 mol O₂ × 32.00 g O₂1 mol O₂ ≈ 47.0 g O₂

Step 2 (Part B): Calculating Percent Yield

Percent Yield = 92.0 g100.0 g × 100 ≈ 92.0%