Calcium carbonate, CaCO₃, decomposes when heated - OCR Gateway - GCSE Chemistry: Combined Science - Question 7 - 2019 - Paper 3

From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CaO. Since 11.2 g of CaO is produced, we calculate the moles of CaO:

The molar mass of CaO is approximately 56.1 g/mol.

Calculating:

$n_{CaO} = \frac{11.2 g}{56.1 g/mol} \approx 0.1994 mol$

Using the stoichiometric ratio from the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, the moles of CO₂ produced is equal to the moles of CaCO₃ used:

$n_{CO₂} = n_{CaCO₃} - n_{CaO} \approx 0.1997 mol - 0.1994 mol = 0.0003 mol$

To find the mass of CO₂, we can use the molar mass of CO₂, which is approximately 44.0 g/mol:

$mass_{CO₂} = n_{CO₂} \times molar\ mass_{CO₂} = 0.0003 mol \times 44.0 g/mol \approx 0.0132 g$

However, if we look at the totals, considering the mass loss from decarbonation:

Total mass of CaCO₃ used = 20.0 g Mass of CaO made = 11.2 g Mass of CO₂ produced:

$mass_{CO₂} = 20.0 g - 11.2 g = 8.8 g$