Determine the limiting reactant, theoretical yield, and percent yield for the reaction.

Question

Consider the reaction between calcium oxide and carbon dioxide:   

                                    CaO (s) + CO2 (g)  → CaCO3 (s) 



A chemist allows 15.27 g of CaO and 18.06 g of CO2 to react. When the reaction is finished, the chemist collects 22.15 g of CaCO3. Determine the limiting reactant, theoretical yield, and percent yield for the reaction. 

Molar masses, g/mol: CaO = 56.08 ; CO2 = 44.01 ; CaCO3 = 100.09 

Answer 1

One way to find the limiting reactant is to divide the moles of each reactant by its coefficient and see which values is less.

CaO: 15.27 g x 1 mol/56.08 g = 0.2723 moles CaO ÷ 1 = 0.2723

CO2: 18.06 g x 1 mol/44.01 g = 0.4104 moles CO2 ÷ 1 = 0.4104

CaO is Limiting

Theoretical yield (determined by the limiting reactant):

0.2723 mol CaO x 1 mol CaCO3 / mol CaO x 100.09 g CaO/mol = 27.25 g CaO theoretical yield

Percent yield = actual yield / theoretical yield (x100%)

% yield = 22.15 g / 27.25 g (x100%) = 81.28% yield