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 14.4 g of CaO and 13.8 g of CO2 to react. When the reaction is finished, the chemist collects 19.4 g of CaCO3. Determine the limiting reactant, theoretical yield, and percent yield for the reaction. 

Molar masses, g/mol: CaO = 56.08 ; CO2 = 28.01 ; CaCO3 = 100.09 

Answer 1

CaO (s) + CO2 (g) → CaCO3 (s)  



Moles of CaO = 14.4 g / 56.08 gmole^-1 = 0.257 moles used

Moles of CO2 = 13.8 g / 28.01 gmole^-1 = 0.492 moles used

Moles of CaCO3 = 19.4g / 100.09 gmole^-1 = 0.194 moles produced.

As the reaction shows that 1 mole of each reactant produces 1 mole if product, CaO is the limiting reactant as it has the fewer number of moles.

Theoretically, 0.257 moles of CaCO3 should be produced.

The percent yield is 0.194 moles CaCO3 / 0.257 moles CaCO3 x 100 = 75.5%

Check the math.