Q.78 Chemistry ⭐ 360° Option Elimination Active

NEET 2023 Chemistry Question 78 Solution

Topic: Some Basic Concepts of Chemistry · Official NTA Booklet Code G5

Question

The right option for the mass of CO₂ produced by heating 20 g of 20% pure limestone is (Atomic mass of Ca = 40) [CaCO₃ → (1200 K) CaO + CO₂]

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Verified Against NTA Final Answer Key

Option (3) : 1.76 g

📚 Some Basic Concepts of Chemistry Medium

📋 Given Parameters & Problem Setup

Sample mass of limestone 20 g

Purity of CaCO₃ 20%

Thermal condition 1200 K

Atomic mass of Ca 40 u

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The 30-Second Exam Hall Shortcut: ⚡ Pure Mass to Mole Direct Ratio: 4 g CaCO₃ → 1.76 g CO₂

⏱️ Save ~40 seconds

Limestone is only 20% pure: Pure CaCO₃ = 20 × 0.20 = 4 g. Since M(CaCO₃) = 100 g/mol, moles = 4/100 = 0.04 mol. Each mole of CaCO₃ yields 1 mole of CO₂ (44 g). Mass of CO₂ = 0.04 × 44 = 1.76 g in 15 seconds!

1 Calculate mass of pure CaCO₃: 20 g × 20% = 4 g.

2 Moles of pure CaCO₃: 4 / 100 = 0.04 mol.

3 Stoichiometry: 1 mol CaCO₃ ⟶ 1 mol CO₂ ⟹ 0.04 mol CO₂ produced.

4 Mass of CO₂: 0.04 mol × 44 g/mol = 1.76 g.

5 Match Option (3) immediately.

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Why You Can 100% Trust This Shortcut: Law of Conservation of Mass & Stoichiometric Mole Ratios in NCERT Chemistry.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determine the Mass of Pure Calcium Carbonate

The limestone sample has a total mass of 20 g and is 20% pure. Only pure CaCO₃ decomposes according to the stoichiometry of the reaction:

\text{Mass of pure } \text{CaCO}_3 = 20\text{ g} \times \frac{20}{100} = 4\text{ g}

Step 2

Calculate Moles of Pure CaCO₃

Compute the molar mass of CaCO₃ and determine moles present in 4 g:

M(\text{CaCO}_3) = 40 + 12 + 3(16) = 100\text{ g mol}^{-1} \implies n(\text{CaCO}_3) = \frac{4\text{ g}}{100\text{ g mol}^{-1}} = 0.04\text{ mol}

Step 3

Apply Reaction Stoichiometry to Find Moles of CO₂

The balanced chemical equation for the thermal decomposition of limestone at 1200 K is:

\text{CaCO}_3(s) \xrightarrow{1200\text{ K}} \text{CaO}(s) + \text{CO}_2(g) \implies n(\text{CO}_2) = n(\text{CaCO}_3) = 0.04\text{ mol}

Step 4

Compute the Mass of Evolved CO₂ Gas

Multiply the moles of CO₂ by its molar mass (M = 12 + 2 × 16 = 44 g/mol):

m(\text{CO}_2) = 0.04\text{ mol} \times 44\text{ g mol}^{-1} = 1.76\text{ g}

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360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 (1)

✗ Incorrect

Verdict & Scientific Reason:

Corresponds to 0.03 mol CO₂ (15% purity assumption).

⚠️

Student Fallacy (Why candidates pick this trap): Candidate miscalculated the percentage purity as 15% instead of 20%.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Corresponds to confusing gas volume at STP (1.12 L) with mass in grams.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate confused the volume of gas at STP with mass in grams.

3 (3)

✓ 100% Correct

Verdict & Scientific Reason:

Accurate: 0.04 mol of pure CaCO₃ thermally decomposes to yield 0.04 mol of CO₂ (0.04 × 44 = 1.76 g).

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Why this is airtight: None. Candidate executed the sample purity calculation correctly.

4 (4)

✗ Incorrect

Verdict & Scientific Reason:

Corresponds to 0.06 mol of CO₂ (30% purity assumption).

⚠️

Student Fallacy (Why candidates pick this trap): Candidate committed an arithmetic error in multiplying moles by molar mass.

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NTA Trap Alert: Ignoring Purity Percentage

The most frequent blunder is ignoring the 20% purity specification and calculating CO₂ directly from 20 g of sample: 20/100 = 0.2 mol → 0.2 × 44 = 8.8 g. Always multiply sample mass by purity fraction first!

🛡️ How to avoid the -1 mark: Write: Pure reactant mass = Total sample mass × (Purity % / 100) before any mole conversions.

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Core NEET Syllabus Standard

National Eligibility cum Entrance Test (NEET UG) · Some Basic Concepts of Chemistry Standard Benchmark

Core Scientific Principle Tested:

"In chemical reactions, stoichiometric coefficients apply exclusively to the mass of pure reacting species."

Syllabus Focus: Stoichiometry and Percentage Purity

Preserved in Rationalized Curriculum Syllabus & Academic Standards ↗

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NEET 2023 Chemistry Question 78 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Pure Mass to Mole Direct Ratio: 4 g CaCO₃ → 1.76 g CO₂

Step-by-Step Rigorous Workflow & Mathematical Proof

Determine the Mass of Pure Calcium Carbonate

Calculate Moles of Pure CaCO₃

Apply Reaction Stoichiometry to Find Moles of CO₂

Compute the Mass of Evolved CO₂ Gas

360° Option Elimination Engine Key Feature

NTA Trap Alert: Ignoring Purity Percentage

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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