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?

The percentage yield of the experiment is 92.4%

The balanced equation for the reaction is given below:

Next, we shall determine the mass of CaO that reacted and the mass of Ca(OH)₂ produced from the balanced equation.

Molar mass of CaO = 40 + 16

= 56 g/mol

Mass of CaO from the balanced equation = 1 × 56 = 56 g

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

From the balanced equation above,

56 g of CaO reacted to produce 74 g of Ca(OH)₂.

Next, we shall determine the theoretical yield of Ca(OH)₂. This can be obtained as follow:

From the balanced equation above,

56 g of CaO reacted to produce 74 g of Ca(OH)₂.

Therefore,

5 g of CaO will react to produce = = 6.61 g of Ca(OH)₂.

Thus, the theoretical yield of Ca(OH)₂ is 6.61 g

Finally, we shall determine the percentage yield of Ca(OH)₂.

Actual yield of Ca(OH)₂ = 6.11 g

Theoretical yield of Ca(OH)₂ = 6.61 g

Percentage yield of Ca(OH)₂ =

Therefore, the percentage yield of the experiment is 92.4%

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explanation:

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol