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 1.00-g sample of CaO is reacted with excess water and 0.82 g of Ca(OH)2 is recovered. What is the percent yield in this experiment?

Answer 1

Answer: The percent yield in this experiment is 63.1 %

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} CaO=(1.00g)/(56g/mol)=0.018moles`

`CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)`

As
`H_2O` is the excess reagent,
`CaO` acts as the limiting reagent and it limits the formation of product.

According to stoichiometry :

1 mole of
`CaO` produce = 1 mole of
`Ca(OH)_2`

Thus 0.018 moles of
`CaO` will produce=
`(1)/(1)* 0.018=0.018moles` of
`Ca(OH)_2`

Mass of
`Ca(OH)_2=moles* {\text {Molar mass}}=0.018moles* 74g/mol=1.3g`

Theoretical yield = 1.3 g

Experimental yield of ethanol = 0.82 g

`\%\text{ yield }=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100=(0.82g)/(1.3g)* 100=63.1\%`

Thus the percent yield in this experiment is 63.1 %