First determine theoretical yield (TY). The actual yield (AY) is given as 1.24g.
mols CaCO3 = grams/molar mass = ?
Look at the equation. You get 1 mol CaO for every 1 mol CaCO3; therefore, mols CaCO3 from above will give you that many mols CaO.
Convert mols CaO to grams. g = mols CaO x molar mass CaO. That is the TY.
Then %yield = (AY/TY)*100 = ?
The decomposition of 3.32 g CaCO3 results in an actual yield of 1.24 g of CaO. What is the percent yield of calcium oxide for this reaction?
CaCO3(s) → CaO(s) + CO2(g)
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