Percentage yield = (actual yield / theoretical yield) x 100%
The balanced equation for the reaction is
CaCO₃(s) → CaO(s) + CO₂(g)
Moles = mass / molar mass
Mass of CaCO₃(s) used = 2.00 x 10³ g
Molar mass of CaCO₃(s) = 100.0 g/mol
Moles of CaCO₃(s) = 2.00 x 10³ g / 100.0 g/mol
= 20.0 mol
Stoichiometric ratio between CaCO₃(s) and CaO(s) is 1 : 1
Hence, moles of CaO = moles of CaCO₃(s)
= 20.0 mol
Molar mass of CaO = 56.0 g/mol
Mass of CaO = 20.0 mol x 56.0 g/mol
= 1120 g
Hence, theoretical yield of CaO = 1120 g
Actual yield = 1.05 x 10³ g = 1050 g
Percentage yield = (1050 g / 1120 g) x 100%
= 93.75%
Hence, percentage yield of CaO is 93.75%
