Limiting Reagent Formula
The reactant that determines maximum product formed - consumed completely first
Concept
In reactions with multiple reactants, the limiting reagent runs out first and limits product formation.
The excess reagent remains after the reaction is complete.
Limiting Reagent
- Consumed completely
- Determines max product
- Controls theoretical yield
- Reaction stops when gone
Excess Reagent
- Some remains unused
- More than needed
- Amount left = initial - consumed
- Can be recovered
Methods to Identify Limiting Reagent
Method 1: Mole Ratio Comparison
Step 1: Convert all reactant masses to moles
Step 2: Divide each by its stoichiometric coefficient
Step 3: Smallest result = limiting reagent
For aA + bB → products:
Compare: (moles A)/a vs (moles B)/b
Method 2: Product Calculation
Step 1: Calculate product from each reactant separately
Step 2: Reactant producing LESS product = limiting reagent
Step 3: Use that amount as theoretical yield
Worked Examples
Example 1: Simple Synthesis
2H₂ + O₂ → 2H₂O
Given: 4.0 g H₂ and 32.0 g O₂
Solution (Method 1):
n(H₂) = 4.0 / 2.02 = 1.98 mol
n(O₂) = 32.0 / 32.00 = 1.00 mol
Divide by coefficients:
H₂: 1.98 / 2 = 0.99
O₂: 1.00 / 1 = 1.00
Limiting Reagent: H₂ (smaller value)
Theoretical yield:
n(H₂O) = 1.98 mol H₂ × (2 mol H₂O / 2 mol H₂) = 1.98 mol
m(H₂O) = 1.98 × 18.02 = 35.7 g
Theoretical Yield: 35.7 g H₂O
Example 2: Excess Calculation
N₂ + 3H₂ → 2NH₃
Given: 14.0 g N₂ and 6.0 g H₂
Solution:
n(N₂) = 14.0 / 28.02 = 0.500 mol
n(H₂) = 6.0 / 2.02 = 2.97 mol
Check ratios:
N₂: 0.500 / 1 = 0.500
H₂: 2.97 / 3 = 0.99
Limiting Reagent: N₂
H₂ consumed:
0.500 mol N₂ × (3 mol H₂ / 1 mol N₂) = 1.50 mol H₂
= 1.50 × 2.02 = 3.03 g H₂
H₂ excess:
6.0 - 3.03 = 2.97 g H₂ remains
Excess H₂: 2.97 g
Example 3: Method 2 (Product Calculation)
4Al + 3O₂ → 2Al₂O₃
Given: 54.0 g Al and 64.0 g O₂
Solution:
n(Al) = 54.0 / 26.98 = 2.00 mol
n(O₂) = 64.0 / 32.00 = 2.00 mol
From Al:
2.00 mol Al × (2 mol Al₂O₃ / 4 mol Al) = 1.00 mol Al₂O₃
From O₂:
2.00 mol O₂ × (2 mol Al₂O₃ / 3 mol O₂) = 1.33 mol Al₂O₃
Limiting Reagent: Al (produces less product)
Theoretical yield: 1.00 mol × 101.96 g/mol = 102 g Al₂O₃
Common Mistakes
Comparing Masses Instead of Moles
Must convert to moles first - can't compare grams directly!
Ignoring Stoichiometric Coefficients
For N₂ + 3H₂ → 2NH₃, need 3× more H₂ moles than N₂
Choosing Largest Value
Limiting reagent gives SMALLEST ratio (n/coefficient), not largest
Base Yield on Limiting Reagent
Always calculate theoretical yield using the limiting reagent amount