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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.

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Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

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.

a. Theoretical yield  for the reaction =26.03 g.

b. Percent yield   for the reaction= 87%

c. Limiting reactant   for the reaction = CaO

Calculation of the theoretical yield, percent yield, and the limiting reactant:

Since

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Now

Chemical equation:

CaO + CO₂   → CaCO₃

Here,

Number of moles of CaO should be

= Mass /molar mass

= 14.4 g / 56.1 g/mol

 = 0.26 mol

Now

Number of moles of CO₂ should be

= Mass /molar mass

= 13.8 g / 44 g/mol

= 0.31 mol

Now

                 CO₂         :                CaCO₃  

                 1               :                 1

                0.31           :              0.31

               CaO           :               CaCO₃  

                1                :                 1

                0.26         :              0.26

Now

1.

Mass of CaCO₃:

Theoretical yield

= moles × molar mass

=0.26 mol × 100.1 g/mol

=  26.03 g

2.

Percent yield:

Percent yield = actual yield / theoretical yield × 100

= 22.6 g/ 26.03 g × 100

= 0.87× 100

= 87%

3.

Limiting reactant:

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

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