Question

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?

Answer 1

Answer:

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 %