A 1.00 g sample containing CaCO3 liberates 0.220 g of CO2 upon treatment with excess HCl. Calculate % CaCO3 and % Ca in the sample.

Answer 1

Molar mass CaCO₃ = 100. g/mol

Molar mass CO₂ = 44.0 g/mol

Atomic mass Ca =40.08 g/mo

CaCO₃ + 2HCl ==> CaCl₂ + CO₂ + H₂O .. balanced equation for reaction taking place

moles of CO₂ produced = 0.220 g CO₂ x 1 mol CO2 / 44.0 g = 5.00x10^-3 moles CO₂

moles CaCO₃ initially present = 5.00x10^-3 moles CO₂ x 1 mol CaCO₃/mol CO₂ = 5.00x10^-3 moles CaCO₃

grams CaCO₃ initially present = 5.00x10^-3 moles CaCO₃ x 100. g/mol = 0.500 g CaCO₃

%CaCO₃ = 0.500 g / 1.00 g (x100%) = 50.0%

%Ca in CaCO₃ = 40.08 g/100. g (x100%) = 40.1%

%Ca in original sample = 40.1% x 50% = 20.1% Ca in original sample