Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO (s) + H₂O (l) → Ca(OH) ₂ (s)
In a particular experiment, a 1.50-g sample of CaO is reacted with excess water and 1.48 g of Ca(OH) ₂ is recovered. What is the percent yield in this experiment?

Question

Quizplus · Accepted Answer

The first step is to calculate the theoretical yield of Ca(OH)2:

Molar mass of CaO = 56.08 g/mol
1.50 g of CaO is equal to 1.50 g / 56.08 g/mol = 0.02675 mol of CaO
The stoichiometry of the balanced equation shows that 1 mol of CaO produces 1 mol of Ca(OH)2
Therefore, the theoretical yield of Ca(OH)2 is also 0.02675 mol

Molar mass of Ca(OH)2 = 74.09 g/mol
0.02675 mol of Ca(OH)2 is equal to 1.98 g of Ca(OH)2
Therefore, the percent yield is:

(percent yield) = (actual yield ÷ theoretical yield) x 100%
(percent yield) = (1.48 g ÷ 1.98 g) x 100% = 74.8%

The correct answer is D, with a percent yield of 74.8%.

Calcium Oxide Reacts with Water in a Combination Reaction to Produce