Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide:
CaO(s)+ $\mathrm { H } _ { 2 }$ O(l)? $\mathrm { Ca } ( \mathrm { OH } ) _ { 2 }$ (s)
In a particular experiment,5.00 g sample of CaO is reacted with excess water and 6.11 g of $\mathrm { Ca } ( \mathrm { OH } ) _ { 2 }$ is recovered.What is the percent yield in this experiment?

Question

Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: 
CaO(s)+ $\mathrm { H } _ { 2 }$ O(l)? $\mathrm { Ca } ( \mathrm { OH } ) _ { 2 }$ (s)
In a particular experiment,5.00 g sample of CaO is reacted with excess water and 6.11 g of $\mathrm { Ca } ( \mathrm { OH } ) _ { 2 }$ is recovered.What is the percent yield in this experiment?

Quizplus · Accepted Answer

The balanced equation tells us that one mole of CaO reacts with one mole of water to produce one mole of Ca(OH)2. We can use the molar mass of Ca(OH)2 to convert the mass recovered to moles of Ca(OH)2 produced:

6.11 g Ca(OH)2 x (1 mol Ca(OH)2 / 74.09 g Ca(OH)2) = 0.0826 mol Ca(OH)2

Since one mole of CaO produces one mole of Ca(OH)2, we know that the theoretical yield of Ca(OH)2 is also 0.0826 mol. We can use the molar mass of CaO to convert the mass used to moles of CaO:

5.00 g CaO x (1 mol CaO / 56.08 g CaO) = 0.0891 mol CaO

The percent yield is then calculated as:

(actual yield / theoretical yield) x 100%
= (0.0826 mol / 0.0891 mol) x 100%
= 92.5%

Calcium Oxide Reacts with Water in a Combination Reaction to Produce