Mole Concept Stoichiometry Question JEE Main 2026

Quick Summary

Question Type: Stoichiometry and mass calculation of product
Chapter: Mole Concept – Stoichiometry
Difficulty: ⭐⭐ Easy
Time to Solve: 3-4 minutes
Key Formula: Moles = mass / molar mass; use balanced equation to convert reactant moles into product moles
Correct Answer: (C) 7.1 g
Why: The balanced reaction is MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O. Molar mass of MnO₂ = 87 g mol⁻¹, so 8.7 g corresponds to 0.1 mol. From the equation, 1 mol MnO₂ gives 1 mol Cl₂. Therefore 0.1 mol Cl₂ is formed, whose mass is 0.1 × 71 = 7.1 g.

The Question

JEE Main 2026 (Online) 21 January Evening Shift – Mole Concept

Aqueous HCl reacts with MnO₂(s) to form MnCl₂(aq), Cl₂(g), and H₂O(l). What is the weight (in g) of Cl₂ liberated when 8.7 g of MnO₂(s) is reacted with excess aqueous HCl solution?

Given: molar mass in g mol⁻¹: Mn = 55, Cl = 35.5, O = 16, H = 1

Choose the correct answer from the options given below:
(A) 14.2
(B) 21.3
(C) 7.1
(D) 71

Quick Answer

Correct Option: (C) 7.1

Reasoning: First write the balanced equation:

• MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O

Now find moles of MnO₂:

• Molar mass of MnO₂ = 55 + 2×16 = 87 g mol⁻¹
• Moles of MnO₂ = 8.7/87 = 0.1 mol

From the balanced equation:

• 1 mol MnO₂ gives 1 mol Cl₂
• So 0.1 mol MnO₂ gives 0.1 mol Cl₂

Mass of Cl₂:

• Molar mass of Cl₂ = 2 × 35.5 = 71 g mol⁻¹
• Mass = 0.1 × 71 = 7.1 g

Therefore, the correct answer is (C) 7.1 g.

Why Other Options Are Wrong

• Option A (14.2): Incorrect because this would correspond to 0.2 mol of Cl₂, but only 0.1 mol MnO₂ is present.
• Option B (21.3): Incorrect because it overestimates the product formed from the given amount of MnO₂.
• Option C (7.1): Correct because 0.1 mol of Cl₂ is formed and its mass is 7.1 g.
• Option D (71): Incorrect because 71 g is the mass of 1 mol of Cl₂, not 0.1 mol.

Video Solution

If you want the full explanation in a clear step-by-step teaching format, watch the video solution below:

Watch Full Video Solution on YouTube

Understanding the Concept

How to Solve Stoichiometry Questions Like This

These questions are straightforward if you follow the correct sequence:

• write the balanced equation
• calculate moles of the given reactant
• use the mole ratio from the equation
• convert final moles into mass

Since HCl is given in excess, MnO₂ becomes the only reactant that controls the amount of Cl₂ formed.

The Key Principle

1. Balanced equation gives mole ratio
2. Given mass must first be converted to moles
3. Excess reactant does not limit product formation
4. Final mass = moles × molar mass

Detailed Step-by-Step Solution

Step 1: Write the Balanced Equation

• MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O

This shows:

• 1 mol MnO₂ gives 1 mol Cl₂

Step 2: Calculate Moles of MnO₂

• Molar mass of MnO₂ = 55 + 2×16 = 87 g mol⁻¹
• Given mass = 8.7 g
• Moles of MnO₂ = 8.7/87 = 0.1 mol

Step 3: Use the Stoichiometric Ratio

Download Handwritten Solution

From the equation:

• 1 mol MnO₂ gives 1 mol Cl₂
• So 0.1 mol MnO₂ gives 0.1 mol Cl₂

Step 4: Convert Cl₂ Moles into Mass

• Molar mass of Cl₂ = 2 × 35.5 = 71 g mol⁻¹
• Mass of Cl₂ = 0.1 × 71 = 7.1 g

Step 5: Write Final Answer

So the correct option is (C).

Final Answer

Option (C): 7.1 ✓

The weight of Cl₂ liberated is 7.1 g.

Essential Formulas for This Topic

Primary Rules

1. Mole Formula:
   • Number of moles = given mass / molar mass
2. Stoichiometric Ratio Rule:
   • Use coefficients from balanced equation to relate reactants and products
3. Mass Formula:
   • Mass = moles × molar mass

Important Notes

• HCl is in excess, so MnO₂ controls the amount of product formed
• 1 mole of MnO₂ gives exactly 1 mole of Cl₂
• Molar mass of Cl₂ must be taken as 71 g mol⁻¹

Common Mistakes to Avoid

❌ Mistake 1: Not Writing the Balanced Equation

Wrong Thinking: “I can directly guess the product mass from the given values.”

Correct Approach: Always begin with the balanced reaction before applying mole ratio.

❌ Mistake 2: Using Wrong Molar Mass of MnO₂

Wrong Thinking: “MnO₂ molar mass is 71 or 87? I’ll approximate.”

Correct Approach: Carefully calculate 55 + 32 = 87 g mol⁻¹.

❌ Mistake 3: Forgetting HCl Is in Excess

Wrong Thinking: “I must calculate moles of both reactants.”

Correct Approach: Since HCl is in excess, MnO₂ alone decides how much Cl₂ forms.

❌ Mistake 4: Taking 71 g as the Final Answer Directly

Wrong Thinking: “Molar mass of Cl₂ is 71, so answer is 71.”

Correct Approach: 71 g is for 1 mole of Cl₂, but only 0.1 mol is formed here.

Key Concept Summary

1. Balance the equation first: It gives the exact mole ratio.
2. Convert mass into moles: 8.7 g MnO₂ = 0.1 mol.
3. Use the reaction ratio: 1 mol MnO₂ gives 1 mol Cl₂.
4. Find product mass: 0.1 mol Cl₂ = 7.1 g.
5. Final answer is 7.1 g: Option (C).

The Golden Rule for Stoichiometry Questions

“Convert the given mass to moles first, then use the balanced equation to find product amount.”

Frequently Asked Questions

Q1: Why is MnO₂ the deciding reactant here?

A: Because HCl is given in excess, so MnO₂ limits the amount of Cl₂ formed.

Q2: What is the molar mass of MnO₂?

A: 87 g mol⁻¹.

Q3: How many moles of Cl₂ are formed?

A: 0.1 mol.

Q4: Why is the final answer not 71 g?

A: Because 71 g is the mass of 1 mole of Cl₂, while only 0.1 mole is formed.

Q5: What is the correct answer?

A: Option (C), 7.1 g.

Prerequisites to Solve This Question

1. Mole concept: Converting mass into moles
2. Balanced equations: Reading stoichiometric coefficients correctly
3. Molar mass calculation: Finding formula mass from atomic masses
4. Stoichiometric mass relation: Converting reactant amount into product amount

After Solving This, You Can:

✅ Solve basic stoichiometry questions confidently
✅ Convert moles into mass correctly
✅ Use balanced equations more effectively
✅ Avoid common Mole Concept calculation mistakes

Study Tips for This Topic

For JEE Main:

1. Always balance the equation first: Many errors begin from wrong stoichiometric ratio.
2. Check which reactant is in excess: That immediately simplifies the problem.
3. Write molar mass separately: It reduces silly calculation mistakes.
4. Keep units clear: Distinguish grams, moles, and molar mass carefully.

Common JEE Variants:

• Mass of gaseous product formed
• Excess reagent based stoichiometry
• Mole to mass conversion questions
• Balanced equation application problems

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐ (2/5) – Easy
JEE Main Frequency: High – Basic stoichiometry is a common part of Mole Concept
JEE Advanced Frequency: Medium – Often combined with limiting reagent or redox concepts
Topic Importance: Very High – Stoichiometry is one of the most fundamental chemistry skills

Written by Nishant Kumar Gupta
Chemistry Educator with 12+ Years of Experience Teaching JEE Aspirants
Founder – PadhoLikhoJEE

Last Updated: April 2026
Question Source: JEE Main 2026 PYQ
Topic: Mole Concept – Stoichiometric Mass Calculation