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

Ans. Moles of CO2 produced = Mass/ Molar mass

                                                = 0.558 g/ (44.00 g/mol)

                                                = 0.012682 mol

Balanced reaction: CaCO3 -----heat---------> CaO + CO2

Stoichiometry: 1 mol CaCO3 produces 1 mol each of CaO and CO2.

So,

Moles of CACO3 in the sample = Moles of CO2 released

                                                = 0.012682 mol

Mass of CaCO3 in sample = Moles x Molar mass

                                                = 0.012682 mol x (100.087 g/mol)

                                                = 1.269 g

% limestone = (Mass of CaCO3 / total mass of sample) x 100

                        = (1.269 g/ 1.506 g) x 100

                        = 84.26 %