Mole Concept Limiting Reagent and Gas Evolution Question JEE Main 2026

Quick Summary

Question Type: Limiting reagent, product mass, and gas evolution statement-based question
Chapter: Mole Concept – Limiting Reagent and Stoichiometry
Difficulty: ⭐⭐⭐ Medium
Time to Solve: 3-4 minutes
Key Formula: Moles = mass / molar mass; at 273 K and 1 atm, 1 mole of gas occupies 22.4 L
Correct Answer: (B) Statement B is incorrect
Why: Ca reacts with excess HCl according to Ca + 2HCl → CaCl₂ + H₂. 14.0 g Ca = 0.35 mol, so 0.35 mol H₂ and 0.35 mol CaCl₂ are formed. Mass of CaCl₂ = 0.35 × 111 = 38.85 g, not 33.3 g. So statement B is the incorrect one.

The Question

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

14.0 g of calcium metal is allowed to react with excess HCl at 1.0 atm pressure and 273 K. Which of the following statements is incorrect?

Given: molar mass in g mol⁻¹ of Ca = 40, Cl = 35.5, H = 1

Choose the correct answer from the options given below:
(A) The limiting reagent is calcium metal.
(B) 33.3 g of CaCl₂ is produced.
(C) 7.84 L of H₂ gas is evolved.
(D) 0.35 mol of H₂ gas is evolved.

Quick Answer

Correct Option: (B) Statement B is incorrect

Reasoning: First write the balanced equation:

• Ca + 2HCl → CaCl₂ + H₂

Now calculate moles of Ca:

• Moles of Ca = 14.0/40 = 0.35 mol

Since HCl is in excess:

• Ca is the limiting reagent
• 0.35 mol of Ca produces 0.35 mol of H₂
• 0.35 mol of Ca produces 0.35 mol of CaCl₂

Now check product mass:

• Molar mass of CaCl₂ = 40 + 2×35.5 = 111 g mol⁻¹
• Mass of CaCl₂ = 0.35 × 111 = 38.85 g

So statement B is incorrect.

Why Other Options Are Correct

• Statement A: Correct because HCl is in excess, so calcium is the limiting reagent.
• Statement C: Correct because 0.35 mol H₂ at 273 K and 1 atm occupies 0.35 × 22.4 = 7.84 L.
• Statement D: Correct because from the balanced equation, 1 mol Ca gives 1 mol H₂, so 0.35 mol Ca gives 0.35 mol H₂.

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 This Type of Statement Question

In such problems, the balanced equation gives everything:

• which reactant is limiting
• how many moles of gas are formed
• how much product is produced

Since HCl is given in excess, calcium completely reacts and directly controls all product calculations.

The Key Principle

1. Convert the given mass into moles
2. Use the balanced equation to find product moles
3. Convert gas moles into volume using 22.4 L at STP-like conditions
4. Check each statement one by one

Detailed Step-by-Step Solution

Step 1: Write the Balanced Equation

• Ca + 2HCl → CaCl₂ + H₂

This shows:

• 1 mol Ca gives 1 mol CaCl₂
• 1 mol Ca gives 1 mol H₂

Step 2: Calculate Moles of Calcium

• Mass of Ca = 14.0 g
• Molar mass of Ca = 40 g mol⁻¹
• Moles of Ca = 14.0/40 = 0.35 mol

Since HCl is in excess, Ca is the limiting reagent.

So statement A is correct.

Step 3: Calculate Moles and Volume of Hydrogen Gas

Download Handwritten Solution

From the equation:

• 1 mol Ca gives 1 mol H₂
• So 0.35 mol Ca gives 0.35 mol H₂

At 273 K and 1 atm:

• 1 mol gas = 22.4 L
• Volume of H₂ = 0.35 × 22.4 = 7.84 L

So:

• Statement C is correct
• Statement D is correct

Step 4: Calculate Mass of CaCl₂

From the equation:

• 1 mol Ca gives 1 mol CaCl₂
• So 0.35 mol Ca gives 0.35 mol CaCl₂

Molar mass of CaCl₂:

• 40 + 2×35.5 = 111 g mol⁻¹

Mass of CaCl₂:

• 0.35 × 111 = 38.85 g

So statement B is incorrect.

Step 5: Write Final Answer

Therefore, the incorrect statement is (B).

Final Answer

Option (B): 33.3 g of CaCl₂ is produced ✗

The incorrect statement is (B), because the correct mass of CaCl₂ formed is 38.85 g.

Essential Formulas for This Topic

Primary Rules

1. Mole Formula:
   • Number of moles = given mass / molar mass
2. Stoichiometric Ratio Rule:
   • Use the balanced equation to relate reactant and product moles
3. Gas Volume Rule:
   • At 273 K and 1 atm, 1 mol gas = 22.4 L

Important Notes

• Excess reagent does not control product formation
• 1 mol Ca gives 1 mol H₂ and 1 mol CaCl₂
• Mass and gas-volume statements should always be checked separately

Common Mistakes to Avoid

❌ Mistake 1: Not Identifying the Limiting Reagent

Wrong Thinking: “Since HCl is not numerically given, I should still compare both reactants.”

Correct Approach: If one reactant is explicitly in excess, the other is automatically the limiting reagent.

❌ Mistake 2: Using the Wrong Molar Mass of CaCl₂

Wrong Thinking: “CaCl₂ mass is close to 100 g mol⁻¹, so approximation is fine.”

Correct Approach: Use exact molar mass: 40 + 2×35.5 = 111 g mol⁻¹.

❌ Mistake 3: Wrong Gas Volume Conversion

Wrong Thinking: “0.35 mol H₂ gives 0.35 L or 3.5 L.”

Correct Approach: Multiply 0.35 by 22.4 L mol⁻¹ to get 7.84 L.

❌ Mistake 4: Missing That the Question Asks for Incorrect Statement

Wrong Thinking: “I found the correct calculations, so I’ll mark the correct statement.”

Correct Approach: Read the question carefully. Here you must identify the incorrect statement.

Key Concept Summary

1. Calcium is limiting: HCl is in excess.
2. 14.0 g Ca = 0.35 mol: This is the key starting point.
3. 0.35 mol H₂ is formed: From 1:1 mole ratio with Ca.
4. Hydrogen volume is 7.84 L: At 273 K and 1 atm.
5. Mass of CaCl₂ is 38.85 g: So statement B is incorrect.

The Golden Rule for Statement-Based Stoichiometry Questions

“Use the balanced equation once, then verify every statement separately.”

Frequently Asked Questions

Q1: Why is calcium the limiting reagent?

A: Because HCl is given in excess, so calcium is completely consumed first.

Q2: How many moles of H₂ are formed?

A: 0.35 mol.

Q3: What is the volume of hydrogen gas evolved?

A: 7.84 L at 273 K and 1 atm.

Q4: What is the correct mass of CaCl₂ formed?

A: 38.85 g.

Q5: Which statement is incorrect?

A: Statement (B).

Prerequisites to Solve This Question

1. Mole concept: Converting mass into moles
2. Balanced equations: Understanding reaction coefficients
3. Molar volume of gas: 22.4 L at 273 K and 1 atm
4. Stoichiometric mass relation: Converting moles into product mass

After Solving This, You Can:

✅ Identify limiting reagent confidently
✅ Calculate product mass correctly
✅ Find gas moles and gas volume from reaction data
✅ Solve incorrect-statement Mole Concept questions faster

Study Tips for This Topic

For JEE Main:

1. Spot the excess reagent quickly: It simplifies the whole question.
2. Keep the 22.4 L value ready: It is used often in gas-volume questions.
3. Check units carefully: Mole, gram, and liter conversions are common traps.
4. Read the final command properly: “incorrect” and “correct” change the answer completely.

Common JEE Variants:

• Mass of salt formed
• Volume of gas evolved
• Limiting reagent identification
• Incorrect statement based stoichiometry questions

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐⭐ (3/5) – Medium
JEE Main Frequency: High – Stoichiometry and gas evolution are common Mole Concept themes
JEE Advanced Frequency: Medium – Often combined with redox or multi-step logic
Topic Importance: Very High – This is a core application of mole concept

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 – Limiting Reagent, Product Mass and Gas Evolution