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 2.50-g sample of CaO is reacted with excess water and 2.80 g of Ca(OH)2 is recovered. What is the percent yield in this experiment

Answer 1

The percent yield of the reaction is 175%.

Step-by-step explanation:

In a chemical reaction, the percent yield is a measure of how efficiently a reaction produces a desired product. It is calculated by dividing the actual yield (the amount of product obtained in the experiment) by the theoretical yield (the amount of product that should have been obtained based on stoichiometry calculations) and multiplying by 100.

In this case, we are given that 2.50 g of CaO reacted to produce 2.80 g of Ca(OH)2. The theoretical yield can be calculated using the stoichiometry of the reaction:

1 mole of CaO reacts with 1 mole of Ca(OH)2. The molar mass of CaO is 56.08 g/mol, so 2.50 g is equal to 0.0446 moles. Therefore, the theoretical yield of Ca(OH)2 is also 0.0446 moles, which is equal to 1.60 g.

The percent yield can now be calculated:

Percent yield = (actual yield / theoretical yield) x 100

= (2.80 g / 1.60 g) x 100

= 175%

Answer 2

The answer to your question is:

% yield = 84.8

Step-by-step explanation:

Data

MW CaO = 56 g

MW H2O = 18 g

MW Ca(OH)2 = 74 g

CaO = 2.5 g

Ca(OH)2 = 2.8 g

Balanced Reaction

CaO + H2O ⇒ Ca(OH)2

56g of CaO --------------- 74 g Ca(OH)2

2.5 g --------------- x

x = 3.3 g of Ca(OH)2

% yield =
`(2.8)/(3.3) x 100`

% yield = 84.8