Calcium carbonate can be decomposed by heating according to the equation Below. WHat is the percentage yield of calcium oxide in this reaction if 12.6g of CaO is obtained when 25.9g of CaCO3 is heated ?
CaCO3(S) -> CaO (s) + CO2 (g)
mols CaCO3 = 25.9/molar mass CaCO3.
mols CaO = same as mols CaCO3 according to the balanced equation.
g CaO = mols CaO x molar mass CaO. This is the theoretical yield(TY). The actual yield (AY) is 12.6g
%yield = (AY/TY)*100 = ?
12.6/25.9*100=48.6
To calculate the percentage yield, we need to compare the actual yield of calcium oxide (CaO) obtained to the theoretical yield. The theoretical yield is the amount of CaO that would be produced if the reaction goes to completion based on the stoichiometry of the balanced equation.
1. Calculate the molar mass of CaO:
- Molar Mass of CaO = 40.08 g/mol (Ca) + 16.00 g/mol (O) = 56.08 g/mol
2. Calculate the moles of CaCO3 used:
- Moles of CaCO3 = Mass of CaCO3 / Molar Mass of CaCO3
- Moles of CaCO3 = 25.9 g / (40.08 g/mol + 12.01 g/mol + 3*16.00 g/mol) = 0.2584 mol
3. From the balanced equation, we can see that 1 mol of CaCO3 produces 1 mol of CaO. Therefore, the theoretical yield of CaO is also 0.2584 mol.
4. Calculate the theoretical mass of CaO:
- Theoretical Mass of CaO = Moles of CaO × Molar Mass of CaO
- Theoretical Mass of CaO = 0.2584 mol × 56.08 g/mol = 14.50 g
5. Calculate the percentage yield:
- Percentage Yield = (Actual Yield / Theoretical Yield) × 100%
- Percentage Yield = (12.6 g / 14.50 g) × 100% = 86.90%
Therefore, the percentage yield of calcium oxide in this reaction is approximately 86.90%.
To find the percentage yield of calcium oxide (CaO) in this reaction, we need to compare the actual yield (12.6 grams) to the theoretical yield, assuming 100% yield.
1. First, determine the molar masses of CaCO3 and CaO:
- CaCO3: Atomic mass of Ca (40.08 g/mol) + Atomic mass of C (12.01 g/mol) + 3 x Atomic mass of O (16.00 g/mol) = 100.09 g/mol
- CaO: Atomic mass of Ca (40.08 g/mol) + Atomic mass of O (16.00 g/mol) = 56.08 g/mol
2. Calculate the theoretical yield of CaO using stoichiometry:
From the balanced equation, we can see that 1 mole of CaCO3 produces 1 mole of CaO.
- Moles of CaCO3 = mass (g) / molar mass (g/mol) = 25.9 g / 100.09 g/mol = 0.259 mol
- Theoretical yield of CaO = moles of CaCO3 x molar mass of CaO = 0.259 mol x 56.08 g/mol = 14.50 g
3. Calculate the percentage yield:
Percentage yield = (Actual yield / Theoretical yield) x 100%
- Actual yield = 12.6 g
- Theoretical yield = 14.50 g
- Percentage yield = (12.6 g / 14.50 g) x 100% = 86.9%
Therefore, the percentage yield of calcium oxide (CaO) in this reaction is approximately 86.9%.