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 4.50-g sample of CaO is reacted with excess water and 5.45 g of Ca (OH) 2 is recovered. What is the percent yield in this experiment?

percentage yield = 91. 60%

Explanation:

The reaction is between calcium oxide and water and the reaction produce calcium hydroxide.

Firstly, we need to write the chemical equation of the reaction and balance the equation.

CaO (s) + H2O (l) →Ca (OH) 2 (s)

Since water is in excess the limiting reactant is CaO and it determines the mass of Calcium hydroxide produced.

Check if the equation is balanced

CaO (s) + H2O (l) → Ca (OH) 2 (s)

molar mass of CaO = 40 + 16 = 56 g

molar mass of Ca (OH) 2 = 40 + 32 + 2 = 74 g

if 56 g of CaO produces 74 g of Ca (OH) 2

4.50 will produce? grams of Ca (OH) 2

cross multiply

theoretical yield of Ca (OH) 2 = (4.5 * 74) / 56

theoretical yield of Ca (OH) 2 = 333 / 56

theoretical yield of Ca (OH) 2 = 5.94642857143

theoretical yield of Ca (OH) 2 = 5.95 g

percentage yield = actual yield/theoretical yield * 100

percentage yield = 5.45/5.95 * 100

percentage yield = 545/5.95

percentage yield = 91.5966386555

percentage yield = 91. 60%