100g of calcium carbonate was decomposed by heating to produce 45g of calcium oxide and carbon dioxide, using the equation given, calculate the Percentage Yield of Calcium Carbonate.
CaCO3 -------> CaO + CO2
Please help me i tried working out but kept on getting stuck after calculating moles for CaCO3 and CaO.
To calculate the percentage yield of calcium carbonate (CaCO3), you need to compare the actual yield of the product (CaO) with the theoretical yield.
Let's break down the steps to solve this problem:
1. Calculate the moles of calcium carbonate (CaCO3):
- The molar mass of CaCO3 is 40.08 g/mol (40.08 g/mol for Ca + 12.01 g/mol for C + 16.00 g/mol for O x 3).
- Divide the given mass of CaCO3 (100 g) by its molar mass to convert grams to moles:
Moles of CaCO3 = 100 g / 40.08 g/mol = 2.497 mol (rounded to three decimal places).
2. Calculate the moles of calcium oxide (CaO):
- The balanced equation tells us that 1 mole of CaCO3 produces 1 mole of CaO. Therefore, the moles of CaO formed will be the same as the moles of CaCO3:
Moles of CaO = 2.497 mol.
3. Calculate the theoretical yield of calcium oxide (CaO):
- The molar mass of CaO is 56.08 g/mol (40.08 g/mol for Ca + 16.00 g/mol for O).
- Multiply the moles of CaO by its molar mass to obtain the theoretical yield in grams:
Theoretical yield of CaO = 2.497 mol × 56.08 g/mol = 140.34 g (rounded to two decimal places).
4. Calculate the percentage yield:
- The percentage yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100:
Percentage Yield = (Actual Yield / Theoretical Yield) × 100.
- The actual yield of CaO is given as 45 g.
- Substitute the values into the formula:
Percentage Yield = (45 g / 140.34 g) × 100 = 32.08% (rounded to two decimal places).
Therefore, the percentage yield of calcium carbonate is approximately 32.08%.
Your post isn't quite clear. You may want percent purity of CaCO3 or you may want percentage yield of the reaction (production of CaO) but to me percentage yield of CaCO3 makes no sense because you aren't making CaCO3. Here is how to do both.
CaCO3 ==> CaO + CO2
mols CaCO3 = grams/molar mass = 100 g/100 = 1 mol
From the equation, 1 mol CaCO3 produces 1 mol CaO; therefore, you will have produced 1 mol CaO. How many grams is that? That's grams CaO = mols CaO x molar mass CaO = 1 x 56 = 56 g CaO produced. That's the theoretical yield. You only produced 45; therefore, the percent yield of CaO(percentyield of the reaction) is
%yield = [(45g CaO/56 g CaO)]*100 = ?. All of that assumes the CaCO3 is 100% to begin with.
Now, suppose the CaCO3 is not 100% initially. You obtained 45 g CaO, that is mols CaO = g/molar mass = 45/56 = approx 0.8 mols CaO. That means you had 0.8 mols CaCO3 initially and that is how many grams? That's g = mols x molar mass = 0.8 x 100 = 80 g pure CaCO3 in that 100 g sample.
%CaCO3 = [(0.8/100)]*100 = ?