Heating 50 g of pure CaCO3 (100 g/mol) gives CaO (56 g/mol) and CO2. If only 22.4 g of CaO is actually obtained, the percent yield is

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Heating $50 g$ of pure $CaCO X_{3}$ ( $100 g/mol$ ) gives $CaO$ ( $56 g/mol$ ) and $CO X_{2}$ . If only $22.4 g$ of $CaO$ is actually obtained, the percent yield is

Theoretical

CaO

= 0.5 mol x 56 = 28 g. Percent yield = actual/theoretical x 100 = 22.4/28 x 100 = 80%.