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 2.00-g sample of CaO is reacted with excess water and 2.14 g of Ca(OH)2 is recovered. What is the percent yield in this experiment? a. 107 b. 1.07 c. 2.88 d. 81.1 e. 93.3

The option closest to the percentage yield is option;

d. 81.1

Explanation:

The given chemical equation of the reaction is presented as follows;

CaO (s) + H₂O (l) → Ca(OH)₂

The mass of CaO in the experiment, m = 2.00 g

The volume of water with which the CaO was reacted = Excess volume of water

Number of moles = Mass/(Molar mass)

The mass of Ca(OH)₂ recovered, actual yield = 2.14 g

The molar mass of CaO = 56.0774 g/mol

The number of moles of CaO in the reaction, n₁ = 2.00 g/(56.0774 g/mol ≈ 0.036 moles

The molar mass of Ca(OH)₂ = 74.093 g/mol

The number of moles of Ca(OH)₂ in the reaction, n₂ = 2.14 g/(74.093 g/mol) ≈ 0.029 moles

From the given chemical reaction, one mole of CaO reacts with one mole of H₂O to produce one mole of Ca(OH)₂

Therefore, 0.036 moles of CaO will produce 0.036 moles of Ca(OH)₂

Mass = Number of moles × Molar mass

The mass of 0.036 moles of Ca(OH)₂ ≈ 0.036 moles × 74.093 g/mol = 2.667348 grams

∴ The theoretical yield of Ca(OH)₂ = 2.667348 grams

The percentage yield = (2.14 g)/(2.667348 grams) × 100 = 80.23%

Therefore, the option which is closest is option d. 81.1.

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !