Question

A sample of impure limestone (calcium carbonate) when heated yields calcium oxide and oxygen gas. A 1.506 g sample of limestone gives 0.558 g of carbon dioxide. This is less than what was expected. Calculate the percent of limestone in the impure sample.

Answer 1

CaCO₃ -----------------> CaO + CO₂

100g 44g

Amount of CaCO₃ = 1.506 g

moles of CaCO₃ = 1.506/1000 = 0.001506 mole = 15.06 mmol

The stoichiometry of the reaction is 1:1 ie one mole of CaCO₃ to produce one mole of CO₂

so 15.06 mmol will product 15.06 mmol of CO₂

Amount of CO₂ = 0.558 = 0.558/44 = 0.01268 mol = 0.01268/1000= 12.68 mmol

Amount of CO₂ produced is 12.268 mmol as compared to the expected 15.06 mmol

So the pecentage of limestone in the impure sample is = (12.6/15.06 ) x 100 = 84.19%

The percent of limestone in the impure sample.is 84.19%