Question

C alcium oxide reacts with water in a combination reaction to produce calcium hydroxide:

2 2 CaO (s) H O (l)Ca(OH) (s) 

In a particular experiment, a 5.00-g sample of CaO is reacted with 125 g of water and 6.11 g of is recovered. 2 Ca(OH)

a. How many moles of each reactant are present?

(Show all work below) (2 pt)

______________ moles CaO ______________ moles H2O

b. Which is the limiting reagent? ______________ is the limiting reagent.

(Show all work below) (2 pt)

c. What is the theoretical yield of calcium hydroxide? ______________ g

(Show all work below) (2 pt)

d. What is the percent yield in this experiment? ______________ %

(Show all work below) (2 pt)

Answer 1

(a)

No. of moles = given mass / Molar mass

No. of moles of CaO = 5.0 / 56.0 = 0.089 Moles

No. of moles of H₂O = 125 / 18 = 6.94

Moles of Ca(OH)₂ = 6.11 / 74.1 = 0.094 moles

(b) CaO + H2O ----> Ca(OH)2

one mole of CaO gives one mole of Ca(OH)₂.

Therefore 0.089 moles of CaO will yield 0.089 moles of Ca(OH)₂.

Mass of Ca(OH) yeilded = Molar mass X No. of moles = 74.1 X 0.089 = 6.59g

hence theoretical yeild = 6.59g

(c) Percental yeild = (Theoretical yeild / experimental yeild) X 100 = 92.7 %