The amount of calcium oxide produced on heating 150 kg limestone (75% pure) is _______kg. (Nearest i — JEE Mains Chemistry Past Papers Chemistry Question
Question
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
💡 Solution & Explanation
**Step 1: Determine the mass of pure CaCO₃** Pure limestone = 150 kg × 75% = 112.5 kg = 112,500 g **Step 2: Calculate molar masses** - CaCO₃ = 40 + 12 + (3 × 16) = 100 g/mol - CaO = 40 + 16 = 56 g/mol **Step 3: Write the decomposition equation** CaCO₃ → CaO + CO₂ **Step 4: Calculate moles of CaCO₃** Moles of CaCO₃ = 112,500 g ÷ 100 g/mol = 1,125 mol **Step 5: Use stoichiometry to find CaO produced** From the equation: 1 mol CaCO₃ produces 1 mol CaO Therefore, moles of CaO = 1,125 mol Mass of CaO = 1,125 mol × 56 g/mol = 63,000 g = 63 kg Therefore, the answer is **63 kg**.