Limiting Reactants
Key Concepts
- Limiting reactant: The reactant that runs out first, limiting the amount of product
- Excess reactant: The reactant left over after the reaction
- Theoretical yield: Maximum product possible based on limiting reactant
- Percent yield: (Actual yield / Theoretical yield) x 100%
Strategy
- Calculate moles of each reactant
- Determine which reactant produces less product (that's the limiting reactant)
- Calculate theoretical yield based on limiting reactant
Practice Problems
Problem 1: For 2H2 + O2 → 2H2O, if you have 4 mol H2 and 1 mol O2, which is limiting?
Need 2 mol H2 per 1 mol O2. Have 4 mol H2, need 2 mol O2 but only have 1. O2 is limiting.
Problem 2: With 1 mol O2 as limiting (above), what's the theoretical yield of H2O?
1 mol O2 → 2 mol H2O. Theoretical yield = 2 mol H2O = 36 g
Problem 3: If actual yield is 30 g H2O and theoretical is 36 g, what's percent yield?
(30/36) x 100% = 83.3%
Problem 4: For N2 + 3H2 → 2NH3, with 2 mol N2 and 4 mol H2, which is limiting?
2 mol N2 needs 6 mol H2 (ratio 1:3). Only have 4 mol H2. H2 is limiting.
Problem 5: Based on #4, how much NH3 can form?
4 mol H2 x (2 mol NH3/3 mol H2) = 2.67 mol NH3
Problem 6: How much excess N2 remains in #4?
4 mol H2 uses 4/3 = 1.33 mol N2. 2 - 1.33 = 0.67 mol N2 excess
Problem 7: Theoretical yield is 50 g, actual is 45 g. Percent yield?
(45/50) x 100% = 90%
Problem 8: If percent yield is 80% and theoretical yield is 100g, what's actual yield?
100g x 0.80 = 80 g
Problem 9: 2Mg + O2 → 2MgO. With 48 g Mg (2 mol) and 16 g O2 (0.5 mol), which limits?
2 mol Mg needs 1 mol O2. Only have 0.5 mol O2. O2 is limiting.
Problem 10: Based on #9, what mass of MgO forms?
0.5 mol O2 → 1 mol MgO = 40 g MgO