Determine which reactant limits the reaction and calculate theoretical yield
Limiting Reactant: The reactant that is completely consumed first, limiting the amount of product formed
Enter coefficients and moles for your balanced equation
1. Calculate moles of product each reactant can produce
2. The reactant that produces less product is limiting
3. Theoretical yield = moles from limiting reactant
4. Excess = initial moles - moles consumed by limiting reactant
The limiting reactant is the substance that is completely consumed first in a chemical reaction, determining the maximum amount of product that can form.
Key Concept:
Think of it like making sandwiches: if you have 10 slices of bread and 3 slices of cheese, cheese is limiting (you can only make 3 sandwiches, with bread left over).
1. Balance the Equation
Ensure stoichiometric coefficients are correct
2. Calculate Product from Each
Determine moles of product from each reactant
3. Identify Limiting Reactant
The one producing less product
4. Calculate Theoretical Yield
Based on limiting reactant only
Problem:
Reaction: 2H₂ + O₂ → 2H₂O
Given: 3.0 mol H₂ and 1.0 mol O₂
Find: Limiting reactant and theoretical yield
Solution:
From H₂:
3.0 mol H₂ × (2 mol H₂O / 2 mol H₂) = 3.0 mol H₂O
From O₂:
1.0 mol O₂ × (2 mol H₂O / 1 mol O₂) = 2.0 mol H₂O
O₂ produces less → O₂ is limiting
Results:
• Limiting reactant: O₂
• Theoretical yield: 2.0 mol H₂O
• Excess H₂: 3.0 - 2.0 = 1.0 mol remaining
Problem:
Reaction: N₂ + 3H₂ → 2NH₃
Given: 2.0 mol N₂ and 5.0 mol H₂
Find: Limiting reactant and theoretical yield
Solution:
From N₂:
2.0 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 4.0 mol NH₃
From H₂:
5.0 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 3.33 mol NH₃
H₂ produces less → H₂ is limiting
Results:
• Limiting reactant: H₂
• Theoretical yield: 3.33 mol NH₃
• Excess N₂: 2.0 - 1.67 = 0.33 mol remaining
| Reaction | Balanced Equation | Stoichiometric Ratio |
|---|---|---|
| Water Formation | 2H₂ + O₂ → 2H₂O | 2 mol H₂ : 1 mol O₂ : 2 mol H₂O |
| Ammonia Synthesis | N₂ + 3H₂ → 2NH₃ | 1 mol N₂ : 3 mol H₂ : 2 mol NH₃ |
| Combustion of Methane | CH₄ + 2O₂ → CO₂ + 2H₂O | 1 mol CH₄ : 2 mol O₂ |
| Iron Oxide Reduction | Fe₂O₃ + 3CO → 2Fe + 3CO₂ | 1 mol Fe₂O₃ : 3 mol CO : 2 mol Fe |
Calculate how much product each reactant can produce. The reactant that produces the least product is limiting.
Divide moles available by stoichiometric coefficient for each reactant. The smallest ratio indicates the limiting reactant.