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Limestone is mostly calcium carbonate, CaCO3, but also contains other minerals. When heated, the CaCO3 decomposes into CaO and CO2. A 1.605 g sample of limestone was heated and gave off 0.657 g of CO2.

(b) Determine the percentage mass of CaCO3 in the limestone.

Answer :

Final answer:

The percentage mass of CaCO₃ in limestone sample can be found by dividing the mass of decomposed CaCO₃ by the original mass of limestone and multiplying by 100. For a sample that released 0.657 g of CO₂ upon heating, the percentage mass of CaCO₃ is calculated to be 93.08%.

Explanation:

To determine the percentage mass of CaCO₃ in limestone, we utilize the mass of CO₂ released upon heating and the molar masses of CaCO₃ and CO₂. The chemical reaction for the decomposition of CaCO₃ is as follows:

CaCO₃ (s) → CaO (s) + CO₂ (g)

Using the molecular weights of CaCO₃ (100.09 g/mol) and CO₂ (44.01 g/mol), first, calculate the moles of CO₂ released by using the mass given:

  • Moles of CO₂ = mass of CO₂ / molar mass of CO₂

Then, because the stoichiometry of the reaction is 1:1, the moles of CO₂ would be equal to the moles of CaCO₃ that decomposed. Calculate the mass of the decomposed CaCO₃:

  • Mass of CaCO₃ = moles of CO₂ × molar mass of CaCO₃

Finally, calculate the percentage mass of CaCO₃ in the original limestone sample:

  • % CaCO₃ = (mass of decomposed CaCO₃ / mass of limestone sample) × 100%

Applying this method to the provided data:

  • Moles of CO₂ = 0.657 g / 44.01 g/mol = 0.01493 mol
  • Mass of CaCO₃ = 0.01493 mol × 100.09 g/mol = 1.494 g
  • % CaCO₃ = (1.494 g / 1.605 g) × 100% = 93.08%

Therefore, the percentage mass of CaCO₃ in the limestone is 93.08%.

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