Limiting Reactant Calculations

The limiting reactant (or limiting reagent) is the reactant that runs out first in a chemical reaction, determining the maximum amount of product that can form. The other reactant(s) are in excess.

Limiting Reactant Method

For each reactant, calculate:

Moles of Product = (moles reactant) × (mole ratio)

Limiting reactant = produces the LEAST product

The Sandwich Analogy

Making sandwiches: 2 slices bread + 1 slice cheese → 1 sandwich

You have: 10 slices bread and 3 slices cheese

  • 10 bread slices → can make 5 sandwiches
  • 3 cheese slices → can make 3 sandwiches
  • Cheese is limiting! You can only make 3 sandwiches
  • Leftover: 4 slices of bread (excess reactant)

Same concept applies to chemical reactions!

Step-by-Step Method

Step 1: Write balanced equation

Make sure coefficients are correct - they give mole ratios

Step 2: Convert to moles

Use n = mass / molar mass for each reactant

Step 3: Calculate product from each reactant

For each reactant, use stoichiometry to find how much product it could make

nproduct = nreactant × (coefficientproduct / coefficientreactant)

Step 4: Identify limiting reactant

Whichever reactant produces the SMALLEST amount of product is limiting

Step 5: Use limiting reactant for final answer

The limiting reactant determines theoretical yield

Detailed Example

Problem: 10.0 g Al reacts with 19.0 g O₂. Which is limiting? How much Al₂O₃ forms?

4 Al + 3 O₂ → 2 Al₂O₃

Molar masses: Al = 27.0 g/mol, O₂ = 32.0 g/mol, Al₂O₃ = 102.0 g/mol

Step 1: Convert to moles

nAl = 10.0 g / 27.0 g/mol = 0.370 mol Al

nO₂ = 19.0 g / 32.0 g/mol = 0.594 mol O₂

Step 2: Calculate Al₂O₃ from Al

Mole ratio: 2 mol Al₂O₃ / 4 mol Al = 1/2

nAl₂O₃ from Al = 0.370 mol Al × (2 mol Al₂O₃ / 4 mol Al)

= 0.185 mol Al₂O₃

Step 3: Calculate Al₂O₃ from O₂

Mole ratio: 2 mol Al₂O₃ / 3 mol O₂ = 2/3

nAl₂O₃ from O₂ = 0.594 mol O₂ × (2 mol Al₂O₃ / 3 mol O₂)

= 0.396 mol Al₂O₃

Step 4: Compare

Al produces: 0.185 mol Al₂O₃ ← SMALLER

O₂ produces: 0.396 mol Al₂O₃

Al is the limiting reactant!

It produces less product, so it runs out first. O₂ is in excess.

Step 5: Calculate mass of product

Use the limiting reactant result:

massAl₂O₃ = 0.185 mol × 102.0 g/mol

= 18.9 g Al₂O₃

Final Answers:

  • Limiting reactant: Al
  • Excess reactant: O₂
  • Theoretical yield: 18.9 g Al₂O₃

Common Mistakes

❌ Comparing grams instead of moles

You can't just compare masses! Must convert to moles and use mole ratios from balanced equation.

❌ Picking the reactant with fewer moles

The reactant with fewer moles isn't always limiting! Must account for stoichiometry. 1 mol of a 1:10 reactant can still be in excess.

❌ Using unbalanced equation

Mole ratios come from coefficients - equation MUST be balanced first!

❌ Using excess reactant for final answer

Always use the limiting reactant to calculate theoretical yield. The excess reactant has leftover!

Related Calculators

Frequently Asked Questions

What is the limiting reactant?

The reactant that gets completely used up first in a reaction, limiting the amount of product that can form. Other reactants are in excess.

How do I find the limiting reactant?

Calculate how much product each reactant could make. Whichever produces the LEAST amount of product is the limiting reactant.

Why can't I just compare moles directly?

Because reactions use reactants in specific ratios! 2H₂ + O₂ → 2H₂O needs 2:1 ratio. Even if you have 10 mol H₂ and 3 mol O₂, you must check stoichiometry.

What happens to the excess reactant?

Some remains unreacted. You can calculate how much was used and subtract from the starting amount to find leftover excess.

Is limiting reactant the same as theoretical yield?

No! Limiting reactant is a reactant. Theoretical yield is the maximum product amount, calculated FROM the limiting reactant.