Quicklime, cao, can be prepared by roasting caco3 to produce ca0 and co2 when 2.00 x 10^3 grams of calcium carbonate are heated, 1.05 x 10^3 grams of calcium oxide is collected. what is the percentage yield?

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%

If there are options, they are b, c, and d. I've done this question in class before :)