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.