Theoretical Yield Example If 4.50 g of HCl are reacted with 15.00 g of CaCO3, according to the following balanced chemical equation, calculate the theoretical yield of CO2. 2HCl + CaCO 3 → CaCl 2 + H 2 O + CO 2 1. Determine the number of moles of one of the products (CO2 in this example) produced if all of each reactant is used up. 4.50 g HCl x 1 mol CO 2 1 mol HCl x = 0.0616 mol CO 2 36.5 g HCl 2 mol HCl 15.00 g CaCO 3 x 1 mol CaCO 3 1 mol CO 2 x = 0.1499 mol CO 2 100.1 g CaCO 3 1 mol CaCO 3 2. Use the smallest number of moles of the product (CO2) from step 1 to calculate the theoretical yield of product (CO2). 0.0616 mol CO 2 x 44.0 g CO 2 = 2.71 g CO 2 1 mol CO 2 Note: Since the reactant, HCl, produces the least amount of product, it is the limiting reactant and the other reactant, CaCO3, is in excess. Percent Yield Example If 2.50 g of CO2 are isolated, after carrying out the above reaction, calculate the percent yield of CO2. 2.50 g CO 2 isolated x 100% = 92.3% yield 2.71 g CO 2 theoretical Notes: If you are given a volume for a reactant, you must determine whether you are working with a pure liquid or a solution. You are probably working with a pure liquid if you are given the starting amount of a reagent in volume without a concentration value. In this case, you need to look up the density; volume substance (mL) x density substance (g/mL) = g substance. Then convert to moles via the MM of the substance. You are definitely working with a solution if you are given the starting amount of a reagent in volume and also a concentration value. a. If the concentration value is molarity (M), volume (L) x M (mol/L) = moles. b. Other concentration definitions must also be used appropriately.