Calcium oxide reacts with water in a combination reaction to produce calcium hyd

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 6.77g of Ca(OH)2 is recovered. What is the percent yield in this experiment? 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 6.77g of Ca(OH)2 is recovered. What is the percent yield in this experiment? 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 6.77g of Ca(OH)2 is recovered. What is the percent yield in this experiment?

Explanation / Answer

CaO(s) + H2O(l) ------------------- Ca(OH)2(s)

mass of CaO = 5.50 grmas

molar mass of CaO = 56.08 gram/mole

number of mole of CaO = 5.50/56.08 =0.09801moles

according to equation

1 mole of Cao = 1 mole of Ca(OH)2

0.0981 moles of CaO = ?

                             = 0.0981x1/1 = 0.0981 moles of Ca(OH)2

number of moles of Ca(OH)2 formed = 0.0981 moles

molar mass of Ca(OH)2 = 74.09 gram/mole

mass of 0.0981 moles of Ca(OH)2 = 0.0981x74.09 =7.268 grams

Theoritical yield of Ca(OH)2 = 7.268 grams.

Actual yield = 6.77 grams

Percent yield = Actual yield/Theoritical yield x100

Percent yield = 6.77/7.268 x100 =93.15%

Percent yield = 93.15%.

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