Theoretcal Yield Example

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 + CaCO3 CaCl 2 + H 2O + 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

1mol HCl 36.5 g HCl

x

1mol CO 2 2 mol HCl

=

0.0616 mol CO 2

15.00 g CaCO3

x 1mol CaCO3 100.1g CaCO3

x 1mol CO 2 1mol CaCO3

= 0.1499 mol CO 2

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 1mol CO 2

= 2.71g 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 2.71g 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.

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