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.

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 %

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