In the following reaction, 28 grams of Calcium Oxide CaO reacts with Carbon Dioxide CO2 to produce Calcium Carbonate CaCO3. How much Calcium Carbonate is produced?
CaO+CO2 ---> CaCO3
CaO + CO2 --> CaCO3
atomic masses
Ca = 40
O = 16
C = 12
so
CaO is 40 + 16 = 56 grams/mol
28/56 = .5 mol
so we get .5 mol of CaCO3
CaCO3 = 40 + 12 + 48 = 100 grams/mol
so we get 50 grams of CaCO3
To determine how much calcium carbonate (CaCO3) is produced, we need to use stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction.
1. Start by writing the balanced chemical equation for the reaction:
CaO + CO2 → CaCO3
2. Determine the molar mass of each substance:
- Molar mass of CaO = 40.08 g/mol
- Molar mass of CO2 = 44.01 g/mol
- Molar mass of CaCO3 = 100.09 g/mol
3. Convert the given mass of CaO to moles:
Moles of CaO = Mass of CaO / Molar mass of CaO
= 28 g / 40.08 g/mol
≈ 0.698 mol (rounded to three decimal places)
4. Use the balanced equation to find the mole ratio between CaO and CaCO3:
From the equation, we see that 1 mole of CaO reacts to produce 1 mole of CaCO3.
5. Calculate the moles of CaCO3 produced:
Moles of CaCO3 = Moles of CaO
= 0.698 mol
6. Convert the moles of CaCO3 back to grams:
Mass of CaCO3 = Moles of CaCO3 × Molar mass of CaCO3
= 0.698 mol × 100.09 g/mol
≈ 69.88 g (rounded to two decimal places)
Therefore, approximately 69.88 grams of calcium carbonate (CaCO3) is produced.