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)CaO(s) + "CO"_2(g)CO2(g) rarr→ "CaCO"_3(s)CaCO3(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"CaO(s)=56.077 g/mol. The molar mass of
"CO"_2(g)"CO2(g) = "44.099g/mol"44.099g/mol
Moles of CaO.
"14.4g CaO(s)"14.4g CaO(s) x "1 mol CaO(s)"/"56.077 g/mol"1 mol CaO(s)56.077 g/mol = "0.257 mol CaO(s)"0.257 mol CaO(s)
Moles of "CO"_2(g)"CO2(g).
"13.8g CO"_2(g)13.8g CO2(g) x "1 mol"/"44.099g/mol"1 mol44.099g/mol = "0.313 mol CO"_20.313 mol CO2
Since there is less "CaO(s)"CaO(s), it is the limiting reactant.
Since the mole ratio of "CaO(s)"CaO(s) and the product "CaCO"_3(s)CaCO3(s) is 1:1, the reaction can produce no more than "0.257 mol CaCO"_3(s)0.257 mol CaCO3(s)
To determine the theoretical yield of "CaCO"_3(s)CaCO3(s) in grams, multiply the number of moles of "CaCO"_3(s)CaCO3(s) times its molar mass.
"0.257 mol CaCO"_3(s)0.257 mol CaCO3(s) x "100.11g"/"1 mol"100.11g1 mol = "25.72g CaCO"_3(s)25.72g CaCO3(s)
The theoretical yield of "CaCO"_3(s)CaCO3(s) in this experiment is "25.72g"25.72g
