Consider the following reaction between calcium oxide and carbon dioxide: CaO (s) + CO2 (g) →CaCO3 (s) A chemist allows 14.4 g of CaO and 13.8 g of CO2 to react. When the reaction is finished, the chemist collects 22.6 g of CaCO3.

a. Determine the theoretical yield for the reaction.

b. Determine the percent yield for the reaction.

c. Determine the limiting reactant for the reaction.

Theoretical yield = 26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given dа ta:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles = Mass / molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass / molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃.

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

CaO : CaCO₃

1 : 1

0.26 : 0.26

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles * molar mass

Mass of CaCO₃ = 0.26 mol * 100.1 g/mol

Mass of CaCO₃ = 26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield * 100

Percent yield = 22.6 g / 26.03 g * 100

Percent yield = 0.87 * 100

Percent yield = 87%

Limiting reactant:

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.