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 5.50-g sample of CaO is reacted with excess water and 6.77 g of Ca (OH) 2 is recovered. What is the percent yield in this experiment?

The % yield is 93.25 %

Explanation:

Step 1: Data given

Mass of CaO = 5.50 grams

Water is in excess

Mass of Ca (OH) 2 produced = 6.77 grams

Molar mass Ca (OH) 2 = 74.09 g/mol

Step 2: The balanced equation

CaO (s) + H2O (l) → Ca (OH) 2 (s)

Step 3: Calculate moles CaO

Moles CaO = mass CaO / molar mass CaO

Moles CaO = 5.50 grams / 56.08 g/mol

Moles CaO = 0.098 moles

Step 4: Calculate moles Ca (OH) 2

For 1 mol CaO we need 1 mol H2O to produce 1 mol Ca (OH) 2

For 0.098 moles CaO we'll have 0.098 moles Ca (OH) 2

Step 5: Calculate mass Ca (OH) 2

Mass Ca (OH) 2 = moles * molar mass

Mass Ca (OH) 2 = 0.098 moles * 74.09 g/mol

Mass Ca (OH) 2 = 7.26 grams

Step 6: Calculate % yield

% yield = (actual yield / theoretical yield) * 100 %

% yield = (6.77 grams / 7.26 grams) * 100%

% yield = 93.25 %

The % yield is 93.25 %