Theoretical Yield

Question

What is the theoretical yield in grams of Ca(OH)₂ if 2.04 g of KOH react with 3.02 g of CaCl₂•2H₂O as shown by the below equation?

CaCl₂(aq) + 2KOH(aq) → Ca(OH)₂(s) + 2KCl(aq)

In the part of the calculation that converts the number of grams to moles for CaCl₂•2H₂O, the two water molecules must be included in the molar mass needed to convert grams to moles (molar mass of CaCl₂•2H₂O is 147.0 g/mol).

Answer 1

CaCl₂(aq) + 2KOH(aq) → Ca(OH)₂(s) + 2KCl(aq)

First, we will find the limiting reactant. The easiest way (IMHO) is to find moles of each reactant and divide each by the corresponding coefficient in the balanced equation.

CaCl₂⋅2H₂O: 3.02 g x 147.0 g/mol = 0.0205 mols (÷1 -> 0.0205)

KOH: 2.04 g x 56.1 g/mol = 0.0364 mols (÷2 -> 0.0182) LIMITING REACTANT

Theoretical yield is determined by the amount of limiting reactant, and is found as follows:

0.0364 mols KOH x 1 mol Ca(OH)₂ / 2 mol KOH x 74.1 g Ca(OH)₂ /molCa(OH)₂ = 1.35 g Ca(OH)₂