You must first be sure the equation is balanced, which it is. Then you must determine the limiting reactant, as that will determine the theoretical yield of calcium carbonate.
#"CaO"(s)# + #"CO"_2(g)# #rarr# #"CaCO"_3(s)#
The mole ratios for this equation are 1:1.
Determine the number of moles of each reactant. Moles are calculated by dividing the mass of each reactant by its molar mass. The molar mass of #"CaO(s)" = "56.077 g/mol"#. The molar mass of
#"CO"_2(g)"# = #"44.099g/mol"#
Moles of CaO.
#"14.4g CaO(s)"# x #"1 mol CaO(s)"/"56.077 g/mol"# = #"0.257 mol CaO(s)"#
Moles of #"CO"_2(g)"#.
#"13.8g CO"_2(g)# x #"1 mol"/"44.099g/mol"# = #"0.313 mol CO"_2#
Since there is less #"CaO(s)"#, it is the limiting reactant.
Since the mole ratio of #"CaO(s)"# and the product #"CaCO"_3(s)# is 1:1, the reaction can produce no more than #"0.257 mol CaCO"_3(s)#
To determine the theoretical yield of #"CaCO"_3(s)# in grams, multiply the number of moles of #"CaCO"_3(s)# times its molar mass.
#"0.257 mol CaCO"_3(s)# x #"100.11g"/"1 mol"# = #"25.72g CaCO"_3(s)#
The theoretical yield of #"CaCO"_3(s)# in this experiment is #"25.72g"#
