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]
- 1.76 g
- 2.64 g
- 1.32 g
- 1.12 g
Answer
(1) 1.76 g
Solution
Mass of pure CaCO3 = 20100 20 = 4 g
Molar mass of CaCO3 = 40 + 12 + 48 = 100 g mol-1
moles of CaCO3 = 4100 = 0.04 mol
1 mol CaCO3 gives 1 mol CO2
mass of CO2 = 0.04 44 = 1.76 g
Answer: (1)