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 = ?
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.
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