The volume in litters of CO2 liberated at STP when 10 grams of 90% pure limestone is heated

Explanation

Here's how to solve the problem:Calculate the amount of pure calcium carbonate (CaCO₂) in the limestone:Mass of limestone = 10 gPurity of limestone = 90%Mass of pure CaCO₃ = 10 g * 0.9 = 9 gSet up the balanced chemical equation for the decomposition of CaCO₃:CaCO₃ (s) → CaO (s) + CO₂ (g)Relate the mass of CaCO₃ to the volume of CO₂ using stoichiometry:From the balanced equation, 1 mole of CaCO₃ decomposes to produce 1 mole of CO₂.Molar mass of CaCO₃ ≈ 100 g/molVolume of 1 mole of CO₂ at STP (standard temperature and pressure) = 22.4 LCalculate the volume of CO₂ liberated:Number of moles of CaCO₃ = Mass of CaCO₃ / Molar mass of CaCO₃ = 9 g / 100 g/mol = 0.09 molVolume of CO₂ liberated = Number of moles of CO₂ * Volume per mole at STP = 0.09 mol * 22.4 L/mol = 2.016 LTherefore, the volume of CO₂ liberated at STP when 10 grams of 90% pure limestone is heated completely is 2.016 liters