After performing the experiment, Calculate the percent yield of this reaction.

Question

An excess of sodium carbonate, Na2CO3, in solution is added to a solution containing 17.85 g CaCl2. After performing the experiment, 12.23 g of calcium carbonate, CaCO3, is produced. Calculate the percent yield of this reaction.

Answer 1

First, write a correctly balanced equation for the reaction taking place:

Na₂CO₃ + CaCl₂ ==> 2NaCl + CaCO₃ ... balanced equation

To find the % yield, we need to know the actual yield (12.23 g) and the theoretical yield (to be calculated)

Theoretical yield. We'll use the molar mass of CaCl₂ to find the moles, and then use the balanced equation and dimensional analysis (unit factor) to get our answer using the molar mass of CaCO₃.

Molar mass CaCl₂ = 111 g / mol

Molar mass CaCO₃ = 100 g / mol

moles CaCl₂ used = 17.85 g x 1 mol / 111 g = 0.1608 moles

theoretical yield: 0.1608 mols CaCl₂ x 1 mol CaCO₃ / 1 mol CaCl₂ x 100 g CaCO₃ / mol = 16.08 g CaCO₃

% yield = actual yield / theoretical yield (x100%) = 12.23 g / 16.08 g (x100%) = 76.06%