3.3 Limiting and Non-Limiting Reactants

Introduction

In a chemical reaction, reactants are not always present in the exact amounts needed to react completely with each other. Often, one reactant is used up before the others. This reactant determines how much product can be formed.

This is where the concepts of limiting reactant and non-limiting reactant (also called excess reactant) become important.

Understanding limiting reactants is essential in:

• Industry — to maximize product yield and minimize waste
• Laboratory work — to predict how much product will form
• Everyday life — to understand reactions like burning fuel or cooking

Definition

Limiting Reactant: A limiting reactant is the reactant that is completely consumed in a reaction. It can also be defined as the reactant that produces the lowest moles of products in a reaction.

Non-Limiting Reactant (Excess Reactant): The reactant in excess is the reactant left unutilized or unreacted after the reaction is complete. It is also referred to as a non-limiting reactant.

In simple words:

• Limiting reactant = the reactant that runs out first and stops the reaction.
• Excess reactant = the reactant that is left over after the reaction stops.

Main Idea

• The limiting reactant determines the maximum amount of product that can be formed.
• The excess reactant is left over after the reaction is complete.
• To identify the limiting reactant, calculate the moles of product formed from each reactant.
• The reactant that produces the least moles of product is the limiting reactant.

Core Concept

Think of it like making sandwiches:

• You have 10 slices of bread and 3 slices of cheese.
• Each sandwich needs 2 slices of bread and 1 slice of cheese.
• You can make only 3 sandwiches because you run out of cheese after 3.
• Cheese is the limiting reactant.
• Bread is the excess reactant (you have 4 slices left over).

In chemistry, the same idea applies:

• The reactant that runs out first limits how much product can be made.
• The reactant that is left over is in excess.

Detailed Explanation

Identification of a Limiting Reactant in a Reaction

A limiting reactant can be recognized by calculating the number of moles of products formed from the given amounts of the reactants, using a balanced chemical equation.

The reactant that produces the least number of moles of products is the limiting reactant.

Example from the Source

Problem: 10 moles of H₂ and 10 moles of O₂ were reacted to produce H₂O. Which one of the reactants is the limiting reactant?

Solution:

The reaction is:

2H_2(g) + O_2(g) → 2H₂O_(g)

From the balanced chemical equation:

2H₂ = 2H₂O

2\ moles of H₂ give 2\ moles of H₂O

10\ moles of H₂ will give 22 × 10 = 10\ moles of H₂O

For O₂:

O₂ = 2H₂O

1\ mole of O₂ gives 2\ moles of H₂O

10\ moles of O₂ = 2 × 10 = 20\ moles of H₂O

Comparison:

• H₂\) produces **10 moles** of \(H₂O
• O₂\) produces **20 moles** of \(H₂O

Conclusion: Since H₂ gives the least number of moles of H₂O (10 moles), H₂ is the limiting reactant.

Worked Example: Zinc and Sulphuric Acid

Problem: 200 g of Zn were reacted with 200 g dil. H₂SO₄. Determine: (a) Limiting reactant (b) Mass of hydrogen produced

Zn_(s) + H₂SO_4(aq) → ZnSO_4(aq) + H_2(g)

Solution:

(a) Limiting Reactant

Given: Mass of Zn = 200\ g Molar mass of Zn = 65.4\ g/mol (source uses 63.5)

No. of moles of Zn = 20065.4 = 3.06\ mol

Mass of H₂SO₄ = 200\ g Molar mass of H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 98\ g/mol

No. of moles of H₂SO₄ = 20098 = 2.04\ mol

From the balanced chemical equation:

1\ mole of Zn produces 1\ mol of H₂

3.06\ mol of Zn = 3.06\ mol of H₂

1\ mol of H₂SO₄ = 1\ mol of H₂

2.04\ moles of H₂SO₄ = 2.04\ mol of H₂

Comparison:

• Zn produces 3.06 moles of H₂
• H₂SO₄\) produces **2.04 moles** of \(H₂

Conclusion: Since H₂SO₄ produces fewer moles of product, H₂SO₄ is the limiting reactant.

(b) Mass of Hydrogen Produced

From the limiting reactant (H₂SO₄):

Moles of H₂ = 2.04\ mol

Mass of H₂ = moles × molar mass

= 2.04 × 2 = 4.08\ g

Final Answer: Mass of hydrogen produced = 4.08 g

Concept Assessment Exercise 3.5

1. Magnesium metal reacts with Sulphur to produce MgS. How many grams of magnesium sulphide (MgS) can be made from 1.50 g of Mg and 1.50 g of sulphur?

Mg_(s) + S_(s) → MgS_(s)

Solution:

Moles of Mg:

n = 1.5024 = 0.0625\ mol

Moles of S:

n = 1.5032 = 0.0469\ mol

From the equation:

1\ mol Mg → 1\ mol MgS

1\ mol S → 1\ mol MgS

Comparison:

• Mg produces 0.0625 mol MgS
• S produces 0.0469 mol MgS

Limiting reactant: S

Mass of MgS:

n = 0.0469\ mol

M(MgS) = 24 + 32 = 56\ g/mol

m = 0.0469 × 56 = 2.63\ g

Answer: 2.63\ g of MgS

2. Zinc and Sulphur react to form Zinc Sulphide:

Zn_(s) + S_(s) → ZnS_(s)

If 10.0 g of Zinc and 10.0 g of Sulphur are available for reaction, determine: (a) The limiting reactant (b) The mass of Zinc Sulphide produced

Solution:

Moles of Zn:

n = 10.065.4 = 0.153\ mol

Moles of S:

n = 10.032 = 0.313\ mol

From the equation:

• 1 mol Zn → 1 mol ZnS
• 1 mol S → 1 mol ZnS

Comparison:

• Zn produces 0.153 mol ZnS
• S produces 0.313 mol ZnS

Limiting reactant: Zn

Mass of ZnS:

M(ZnS) = 65.4 + 32 = 97.4\ g/mol

m = 0.153 × 97.4 = 14.9\ g

Answer: (a) Limiting reactant = Zn (b) Mass of ZnS = 14.9 g

3. Aluminium reacts with bromine to form Aluminium bromide:

2Al_(s) + 3Br_2(l) → 2AlBr_3(s)

If 54 g of Al and 200 g of Br₂ are available for reaction, determine: (a) The limiting reactant (b) The mass of AlBr₃ produced

Solution:

Moles of Al:

n = 5427 = 2.0\ mol

Moles of Br₂:

n = 200160 = 1.25\ mol

From the equation:

2\ mol Al → 2\ mol AlBr₃

2.0\ mol Al → 2.0\ mol AlBr₃

3\ mol Br₂ → 2\ mol AlBr₃

1.25\ mol Br₂ → 23 × 1.25 = 0.833\ mol AlBr₃

Comparison:

• Al produces 2.0 mol AlBr₃
• Br₂ produces 0.833 mol AlBr₃

Limiting reactant: Br₂

Mass of AlBr₃:

M(AlBr₃) = 27 + (3 × 80) = 27 + 240 = 267\ g/mol

m = 0.833 × 267 = 222.4\ g

Answer: (a) Limiting reactant = Br₂ (b) Mass of AlBr₃ = 222.4 g

Summary Table

Intermediate Level

Steps to identify the limiting reactant:

• Write the balanced chemical equation.
• Convert the given masses (or moles) of each reactant to moles.
• Use the mole ratio from the equation to calculate the moles of product formed from each reactant.
• The reactant that produces the least moles of product is the limiting reactant.
• Use the limiting reactant to calculate the mass of product.

Key point: The limiting reactant determines the theoretical yield of the reaction.

Important Points

• - Limiting reactant = completely consumed in a reaction; produces the least moles of product.
• - Excess reactant = left unreacted after the reaction is complete.
• - The limiting reactant determines the maximum amount of product.
• - To identify the limiting reactant, calculate the moles of product formed from each reactant.
• - The reactant that produces the least moles of product is the limiting reactant.
• - The limiting reactant is not always the one with the smaller mass or fewer moles.
• - Use the limiting reactant to calculate the mass of product.

Remember

• - Limiting reactant = runs out first.
• - Excess reactant = left over.
• - Limiting reactant determines product amount.
• - Calculate moles of product from each reactant.
• - Least moles of product → limiting reactant.
• - Limiting reactant → theoretical yield.
• - Not always the smaller mass or fewer moles.

Common Mistakes

Quick Revision

Limiting Reactant: A limiting reactant is the reactant that is completely consumed in a reaction. It can also be defined as the reactant that produces the lowest moles of products in a reaction.

• Excess Reactant (Non-Limiting Reactant):
• The reactant in excess is the reactant left unutilized or unreacted after the reaction is complete.

Identification of a Limiting Reactant:

• Calculate the number of moles of products formed from the given amounts of the reactants, using a balanced chemical equation.
• The reactant that produces the least number of moles of products is the limiting reactant.

• Example:
• 10 moles of H₂ and 10 moles of O₂ react to produce H₂O:

• 2H₂ + O₂ → 2H₂O

• H₂\) produces 10 moles of \(H₂O
• O₂\) produces 20 moles of \(H₂O
• H₂ is the limiting reactant.

• Worked Example:
• 200 g Zn + 200 g H₂SO₄:

• Zn + H₂SO₄ → ZnSO₄ + H₂

• Moles of Zn = 20065.4 = 3.06\ mol
• Moles of H₂SO₄ = 20098 = 2.04\ mol
• Zn produces 3.06 mol H₂
• H₂SO₄\) produces 2.04 mol \(H₂
• H₂SO₄ is the limiting reactant.
• Mass of H₂ = 2.04 × 2 = 4.08\ g

60-Second Revision

• Limiting reactant = completely consumed; produces least moles of product.
• Excess reactant = left over after reaction.
• Limiting reactant determines maximum product.
• To find it: calculate moles of product from each reactant.
• Least moles of product → limiting reactant.
• Not always smaller mass or fewer moles.
• Depends on mole ratio from balanced equation.
• Limiting reactant → theoretical yield.
• ⚠️ Warning: always balance the equation before comparing mole ratios.