Question

consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of cao and 13.8 g of co 2 to react. when the reaction is finished, the chemist collects 19.4 g of caco 3 . determine the limiting reactant, theoretical yield, and percent yield for the reaction.

Answer 1

Step-by-step explanation:

Given data:

Mass of calcium oxide = 14.4 g

Mass of carbon dioxide = 13.8 g

Actual yield of calcium carbonate = 19.4 g

Mass of calcium carbonate produced = ?

Limiting reactant = ?

Percent yield = ?

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles of CaO = Mass /molar mass

Number of moles of CaO = 14.4 g / 56.1g/mol

Number of moles of CaO = 0.26 mol

Number of moles of CO₂:

Number of moles of CO₂= Mass /molar mass

Number of moles of CO₂ = 13.8 g / 44 g/mol

Number of moles of CO₂ = 0.31 mol

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

CaO : CaCO₃

1 : 1

0.26 : 0.26

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

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

Limiting reactant:

CaO

Theoretical yield:

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ = 0.26 mol × 100 g/mol

Mass of CaCO₃ = 26 g

Percent yield:

Percent yield = Actual yield / theoretical yield × 100

Percent yield = 19.4 g/ 26 g× 100

Percent yield = 74.6 %