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

Chemistry ma’am · 4 steps

Only the pure part decomposes, so take 75% of the limestone first, then use the 100 to 56 mass ratio.

1. Pure limestone. Only the CaCO₃ in the sample reacts. Its mass is 0.75 × 150 = 112.5 kg. The other 25% is impurity and gives no CaO.

   CaCO₃ →(Δ) CaO + CO₂

2. Mass ratio. Molar masses: CaCO₃ = 40 + 12 + 48 = 100 g mol⁻¹ and CaO = 40 + 16 = 56 g mol⁻¹. One mole gives one mole, so 100 kg of CaCO₃ gives 56 kg of CaO.
3. Put in numbers. 112.5 kg is (112.5 × 10³)/100 = 1125 mol of CaCO₃. That gives 1125 mol of CaO, which is 1125 × 56 = 63000 g.
4. So. The mass of calcium oxide produced is 63 kg, so the answer is 63.

The common slip: 84. Skipping the 75% purity and using all 150 kg as CaCO₃ gives 150 × 56/100 = 84, not 63.

The idea to keep. With an impure sample, convert to the mass of the pure substance before any mole calculation. A 1 : 1 mole ratio then turns into a simple ratio of molar masses.

Answer from the NTA final answer key. Solution by PrepWiser.