Limiting Reagent : Examples & MCQs with Solutions for JEE, NEET, Class 11

ANAND CLASSES Study Material and Notes to learn the concept of Limiting Reagent with easy-to-understand examples and solved MCQs. Perfect for JEE, NEET, and Class 11 Chemistry students. Includes detailed explanations for conceptual clarity.

🔥 Limiting Reagent – Detailed Concept with Basic Examples

🔷 What is a Chemical Reaction? (Before understanding limiting reagent)

A chemical reaction involves reactants turning into products.

🧪 Example: $$\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$$

In this reaction:

• Hydrogen (H₂) and Chlorine (Cl₂) are reactants
• Hydrogen chloride (HCl) is the product

But in real life, you may not always mix the exact amount of H₂ and Cl₂ as needed in the balanced equation. So, what happens then?

That brings us to the concept of the limiting reagent.

🌟 What is a Limiting Reagent?

✅ Definition:

The limiting reagent (or limiting reactant) is the reactant that gets used up first in a chemical reaction, stopping the reaction from continuing and limiting the amount of product formed.

🧪 Key Point:
The limiting reagent determines the maximum amount of product that can be formed in a reaction.

🔍 Why do we call it ‘limiting’?

Because it limits or controls how much product can be made.
Even if other reactants are present in excess, the reaction cannot proceed once the limiting reagent is consumed.

🔄 What about the other reactants?

Those are called excess reagents — they are not fully used up.

📘 Basic Analogy (Daily Life Example):

Imagine you’re making sandwiches 🥪
Each sandwich requires:

• 2 slices of bread
• 1 slice of cheese

You have:

• 10 slices of bread
• 3 slices of cheese

👉 How many sandwiches can you make?

Let’s calculate:

• 10 slices of bread make 5 sandwiches (because 2 per sandwich)
• 3 slices of cheese make 3 sandwiches

So, even though you have enough bread for 5 sandwiches, you can only make 3 sandwiches because cheese runs out first.

✅ Cheese is the limiting reagent.
✅ Bread is the excess reagent.

What is the importance of identifying the limiting reagent?

Answer:

1. It allows us to accurately calculate the amount of product that can be formed.
2. It helps in preventing wastage of excess reactants.
3. It is essential in industrial chemistry for cost-effective production.

How do we determine the Limiting Reagent?

Answer:
Follow these steps:

Step 1: Write the balanced chemical equation.

Make sure the equation is balanced with correct stoichiometric coefficients.

Step 2: Convert the given quantities (mass or volume) into moles of each reactant.

Step 3: Divide the number of moles of each reactant by their respective coefficients in the balanced equation.

$$\text{Ratio} = \frac{\text{Given moles}}{\text{Stoichiometric coefficient}}​$$

Step 4: Compare the ratios.

The smallest ratio indicates the limiting reagent.

Provide a solved example of identifying the limiting reagent

Answer:

Example:
For the reaction: $$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$$

Suppose we have:

• 28 g of N₂
• 6 g of H₂

🔹 Step 1: Calculate moles

• Molar mass of N₂​ = 28 g/mol → Moles = 28/28 = 1 mol
• Molar mass of H₂ = 2 g/mol → Moles = 6/2 = 3 mol

🔹 Step 2: Use the mole ratio from the balanced equation
1 mole of N₂​ reacts with 3 moles of H₂

So we have:

• Required: 1 mol N₂​ : 3 mol H₂
• Given: 1 mol N₂​ and 3 mol H₂

✅ Hence, both are in exact proportion — no limiting reagent.
But now, let’s change the amount of hydrogen to 2 moles.

Now:

• $\frac{1}{1}$ = 1 (for N₂)
• $\frac{2}{3} $= 0.6632​=0.66 (for H₂)

Since 0.66 is less than 1,
🟩 Hydrogen (H₂) is the limiting reagent.

What is an Excess Reagent?

Answer:
An excess reagent is a reactant that is not completely used up in the reaction. It is present in more than the required amount, based on stoichiometry.

🔄 The excess reagent remains unreacted after the completion of the reaction.

What is the relation between Limiting Reagent and Product Formed?

Answer:
The limiting reagent controls the amount of product formed. Once it is consumed, the reaction stops.

Product formed ∝ Amount of limiting reagent

So, to calculate the theoretical yield of the product, we use the moles of the limiting reagent.

🧪 Simple Chemical Example 1:

Let’s look at the reaction: $$\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$$

Suppose we have:

• 1 mole of H₂
• 2 moles of Cl₂

➤ Step-by-step:

1. Balanced Equation:
   $$\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$$
   The mole ratio is 1:1
2. Given moles:
   • H₂ = 1 mole
   • Cl₂ = 2 moles
3. According to the balanced equation, 1 mole of H₂ needs 1 mole of Cl₂
   So 1 mole of H₂ will react with 1 mole of Cl₂ to form 2 moles of HCl.

👉 What’s left?

• H₂ is completely used up
• Cl₂ still has 1 mole left → excess

✅ So, Hydrogen (H₂) is the limiting reagent

✅ Product formed = 2 moles of HCl

🧪 Simple Chemical Example 2:

$$\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$

Suppose:

• H₂ = 3 moles
• O₂ = 2 moles

➤ Step-by-step:

1. Balanced equation mole ratio:
   H₂ : O₂ = 2 : 1
2. Convert to a common scale:
   • Available H₂ / required H₂ = 3 / 2 = 1.5
   • Available O₂ / required O₂ = 2 / 1 = 2
3. The smaller value (1.5) tells us that H₂ is the limiting reagent

✅ So, H₂ will run out first and limit the reaction.

🔬 Why Do We Use Moles?

Because atoms and molecules are so small, we can’t count them individually.
So we use moles, a unit that helps us count molecules in large numbers.

1 mole = 6.022×10²³ particles

We always convert grams to moles when finding the limiting reagent.

🧮 Mass-Based Example:

Let’s use this reaction: $$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$$

Given:

• 28 g of N₂
• 6 g of H₂

Step 1: Calculate moles

• Molar mass of N₂ = 28 g/mol → moles = 28/28 = 1 mol
• Molar mass of H₂ = 2 g/mol → moles = 6/2 = 3 mol

Step 2: Use mole ratio

From the balanced equation:

• 1 mole N₂ reacts with 3 moles H₂

You have exactly:

• 1 mole N₂
• 3 moles H₂

➡️ So both are in the exact ratio. No limiting reagent here.

Now suppose you had:

• 28 g N₂ (1 mole)
• 4 g H₂ (2 moles)

Then:

• N₂: 1 mole
• H₂: 4 g / 2 = 2 moles

Expected ratio: N₂ : H₂ = 1 : 3
Given: 1 : 2

✅ H₂ is less than needed → it’s the limiting reagent

📝 Key Points to Remember

Can Limiting Reagent be identified directly from mass?

Answer:
No. You must convert mass to moles and then use the mole ratio to determine the limiting reagent. Mass alone can be misleading without stoichiometric analysis.

Common Mistakes to Avoid While Identifying Limiting Reagent

⚡ Quick Recap

• Limiting Reagent = reactant used up first
• Excess Reagent = leftover reactant
• Always use moles, not just mass
• Compare mole ratios to identify limiting reagent
• Product formed depends on the limiting reagent only

🧠 Conceptual MCQs on Limiting Reagent

Q1. In the reaction:

$$\text{C} + \text{O}_2 \rightarrow \text{CO}_2$$

If 12 g of carbon (C) reacts with 32 g of oxygen (O₂), what is the limiting reagent?

A. Carbon (C)
B. Oxygen (O₂)
C. Carbon dioxide (CO₂)
D. Both are in the exact ratio (no limiting reagent)

✅ Solution:

• Molar mass of C = 12 g/mol → moles of C = 12/12 = 1 mol
• Molar mass of O₂ = 32 g/mol → moles of O₂ = 32/32 = 1 mol

From the balanced equation: $$\text{C} + \text{O}_2 \rightarrow \text{CO}_2$$

Mole ratio = 1 : 1

Given: 1 mole C and 1 mole O₂ → exact ratio.

🟩 Correct Answer: D. Both are in the exact ratio (no limiting reagent)

Q2. In the reaction:

$$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$$

You have 1 mole of nitrogen gas and 2 moles of hydrogen gas. What is the limiting reagent?

A. Nitrogen
B. Hydrogen
C. Ammonia
D. None of these

✅ Solution:

From the balanced equation:

• 1 mol N₂ reacts with 3 mol H₂

Given:

• 1 mol N₂
• Only 2 mol H₂ (less than 3 mol needed)

🟩 Hydrogen is less than required → it will be used up first

🟩 Correct Answer: B. Hydrogen

Q3. In a reaction:

$$\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3$$

If you start with 4 moles of Al and 6 moles of Cl₂, what is the limiting reagent?

A. Aluminium (Al)
B. Chlorine (Cl₂)
C. Both are limiting
D. No limiting reagent

✅ Solution:

Balanced equation mole ratio:
Al : Cl₂ = 2 : 3

Let’s divide each given mole by its stoichiometric coefficient:

• Al: 4/2 = 2
• Cl₂: 6/3 = 2

👉 Both give the same value = 2 → exact ratio

🟩 Correct Answer: D. No limiting reagent

Q4. 10 g of H₂ and 80 g of O₂ are mixed and reacted. What is the limiting reagent in the formation of water?

$$\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$

A. Hydrogen
B. Oxygen
C. Water
D. Cannot be determined

✅ Solution:

Molar masses:

• H₂ = 2 g/mol → moles = 10/2 = 5 mol
• O₂ = 32 g/mol → moles = 80/32 = 2.5 mol

Mole ratio needed = 2:1 (H₂:O₂)

• Available ratio = 5:2.5 = 2:1 → exact stoichiometric ratio

🟩 So both reactants are in the correct ratio

🟩 Correct Answer: D. Cannot be determined (based on the trick in question, but technically both are in correct ratio, so no limiting reagent)

Q5. Which of the following statements is true about the limiting reagent?

A. It is the reactant that is present in the greatest amount
B. It is the reactant that determines the amount of product formed
C. It is always the one with the smallest mass
D. It is the reactant that remains after the reaction is complete

✅ Solution:

• Limiting reagent determines the maximum product that can be formed.
• It may or may not be the one in smallest mass.
• It is used up first, so it doesn’t remain.

🟩 Correct Answer: B. It is the reactant that determines the amount of product formed

Q6. A reaction requires 1 mol of A and 2 mol of B to make product C. If you have 5 mol of A and 5 mol of B, what is the limiting reagent?

A. A
B. B
C. C
D. No limiting reagent

✅ Solution:

Required ratio A : B = 1 : 2
Available = 5 : 5

Now divide each by their respective coefficients:

• A: 5/1 = 5
• B: 5/2 = 2.5

Smaller value = 2.5 (from B) → B will be used up first

🟩 Correct Answer: B. B