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%

since electrons are negative the answer is neagtaive