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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metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

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