A 18.7 g sample of CaCO3 was treated with aqueous H2SO4 producing calcium sulfate, water and 3.65 g of CO2(g) according to:

Question

CaCO3(s) + H2SO4(aq) -> CaSO4(s) + H2O + CO2(g).
What was the percent yield of CO2?

Answer 1

mols CaCO3 = grams/molar mass = ?
Using the coefficients in the balanced equation convert mols CaCO3 to mols CO2.
Now convert mols CO2 to grams CO2. g = mols x molar mass = ? This is the theoretical yield (TY). The actual yield (AY) in the problem is 3.65 grams.
%yield = (AY/TY)*100 = ?

Answer 2

44.4