Limestone Heating Numerical – JEE Main PYQs | Some Basic Concepts of Chemistry

The amount of calcium oxide produced on heating 150 kg limestone (75% pure) is ______ kg. (Nearest integer)

Given: Molar mass (in g mol⁻¹) of Ca = 40, O = 16, C = 12

Step 1: Write the chemical reaction

CaCO₃ → CaO + CO₂

Step 2: Calculate mass of pure calcium carbonate

Given limestone is 75% pure.
Pure CaCO₃ = 150 × 0.75 = 112.5 kg

Step 3: Calculate molar masses

Molar mass of CaCO₃ = 40 + 12 + (3 × 16) = 100 g mol⁻¹
Molar mass of CaO = 40 + 16 = 56 g mol⁻¹

Step 4: Use stoichiometric relation

100 g of CaCO₃ produces 56 g of CaO

Step 5: Calculate amount of CaO formed

CaO produced = (56 / 100) × 112.5
= 63 kg

✅ Final Answer: 63 kg
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Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.

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