Question

Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide. Ca) + H2O --> Ca(OH)2 In a particular experiment, a 5.00 g sample of CaO is reacted with excess water and 6.11 g of Ca(OH)2 is recovered. What is the percent yield in this experiment?

Answer 1

Answer:

Percent yield = 92.5%

Explanation:

The question asks for the percent yield which can be defined as:

$\frac{actual yield}{theoretical yield} .100$

Where the actual yield is how much product was obtained, in this case 6.11 g of Ca(OH)₂, and the theoretical yield is how much product could be obtained with the given reactants theoretically, that is if the reaction would work perfectly. So we need to calculate first the theoretical yield.

1. First lets write the chemical equation reaction correctly and check that it is balanced:

CaO + H₂O → Ca(OH)₂

2. Calculate the amount of product Ca(OH)₂ that can be obtained with the given reactants (theoretical yield), which are 5.00g of CaO and excess of water. So the amount of CaO will determined how much Ca(OH)₂ we can obtained.

For this we'll use the molar ratio between CaO and Ca(OH)₂ which we see it is 1:1. For every mol of CaO we'll obtain a mol of Ca(OH)₂. So lets convert the 5.00 g of CaO to moles:

 Molar Mass of CaO: 40.078 + 15.999 = 56.077 g/mol

 moles of CaO = 5.00 g / 56.077 g/mol = 0.08916 moles

As we said before from the molar ratio moles of Ca(OH)₂ = moles of CaO

So the moles of Ca(OH)₂ that can be obtained are 56.077 g/mol

We need to convert this value to grams:

 Molar Mass of Ca(OH)₂ = 40.078 + (15.999 + 1.008)*2 = 74.092 g/mol

Theoretical yield of Ca(OH)₂ = 0.08916 moles x 74.092 g/mol = 6.606 g

3. Calculate the percent yield:

$\frac{actual yield}{theoretical yield} .100$

Percent yield = (6.11 g / 6.606g) x 100 = 92.5 %

Answer 2

Answer: 89%

Explanation:

$CaO+H_2O\rightarrow Ca(OH)_2$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of calcium oxide}=\frac{5.0g}{56g/mol}=0.09moles$

$\text{Number of moles of calcium hydroxide acually produced}=\frac{6.11g}{74g/mol}=0.08moles$

According to stoichiometry:

1 mole of $CaO$ gives 1 mole of $Ca(OH)_2$

Thus 0.09 moles of $CaO$ will give=$\frac{1}{1}\times 0.09=0.09moles$ of $Ca(OH)_2$

Thus percentage yield = $\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100}=\frac{0.08}{0.09}\times 100=89\%$

Therefore, the percentage yield in the experiment is 89%.