Question

Dissolving 3.82 g of an impure sample of calcium carbonate in hydrochloric acid produced 0.881 L of carbon dioxide (measured at 20.0°C and 737 mmHg). Calculate the percent by mass of calcium carbonate in the sample.

Answer 1

Balanced reaction : CaCO₃(s) + 2 HCl(aq) CaCl₂(aq) + CO₂(g) + H₂O(l)

volume of CO₂ produced = 0.881 L

pressure of CO₂ = 737 mmHg

pressure of CO₂ = 737 mmHg * (1 atm / 760 mmHg)

pressure of CO₂ = 0.970 atm

temperature of CO₂ = 20.0 ^oC = (20.0 + 273) K = 293 K

According to ideal gas law,

moles CO₂ produced = [(pressure of CO₂) * (volume of CO₂ produced)] / [(R) * (temperature of CO₂)]

where R = 0.0821 L-atm/mol-K

moles CO₂ produced = [(0.970 atm) * (0.881 L)] / [(0.0821 L-atm/mol-K) * (293 K)]

moles CO₂ produced = 0.0355 mol

moles CaCO₃ in sample = moles CO₂ produced

moles CaCO₃ in sample = 0.0355 mol

mass CaCO₃ in sample = (moles CaCO₃ in sample) * (molar mass CaCO₃)

mass CaCO₃ in sample = (0.0355 mol) * (100.0869 g/mol)

mass CaCO₃ in sample = 3.55 g

mass % CaCO₃ in sample = (mass CaCO₃ in sample / mass of sample) * 100

mass % CaCO₃ in sample = (3.55 g / 3.82 g) * 100

mass % CaCO₃ in sample = (0.931) * 100

mass % CaCO₃ in sample = 93.1 %