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 2.00-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

Answer 1

Answer:

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

$Percentage \ yield = \dfrac{Actual \ yield}{Theoretical \ yield} \times 100 \%$

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

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