So, start with the balanced chemical equation for the decomposition of #CaCO_3#
#CaCO_(3(s)) -> CaO_((s)) + CO_(2(g))#
Notice that you've got #"1:1"#mole ratios between all the compounds involved in the reaction. This is important because you'll use the mole ratio #CaCO_3# has with #CaO# to determine what the theoretical yield of the reaction is.
You can double check the result by using moles instead of grams. You know that you have #"20 moles"# of #CaO# produced in theory. The actual number of moles of #CaO# produced is