First there is no such thing as calcium(iii) oxycarbonate. Ditto for calcium(iv) oxide.
CaCO3 + 2HCl ==> H2O + CO2 + CaCl2
mols CaCO3 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols CaCO3 to mols CaO.
Convert mols CaO to grams CaO by grams = mols x molar mass = ?. This is the theoretical yield (TY). The actual yield (AY) given in the problem is 3.65.
%yield = (AY/TY)*100 = ?
Calcium carbonate was reacted with hydrogen chloride. What is the percentage yield of carbon(iv)oxide produced if 3.65g of the gas is collected when 10g of calcium(iii)oxocarbonate reacts
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