The combustion of methane(CH4) produces carbon dioxide and water. Assume that 2.0 mol of CH4 burned in the presence of excess air. What is the percentage yield if in an experiment the reaction produces 87.0 g of CO2?

The answer to your question is 98.9 %

Explanation:

Data

moles of methane = CH₄ = 2.0

excess air

Percent yield = ?

mass of CO₂ = 87 g

- Balanced chemical reaction

                CH₄  +  2O₂   ⇒  CO₂  +  2H₂O

      Reactants     Elements       Products

             1                    C                   1

             4                   H                   2

             4                   O                   2

- Calculate the molar mass of CH₄

CH₄ = 12 + 4 = 16 g

- Convert the moles to mass

                     16 g of CH₄ 1 mol

                       x                  2 moles

                       x = (2 x 16) / 1

                       x = 32 g of CH₄

-Calculate the theoretical formation of CO₂

                     16 g of CH₄ 44 g of CO₂

                     32 g of CH₄  x

                            x = (32 x 44) / 16

                            x = 88 g of CO₂

-Calculate the Percent yield

     Percent yield = Actual yield/Theoretical yield x 100

     Percent yield = 87/88 x 100

    Percent yield = 98.9 %

98.9 %m is the correct answer.

Explanation:

the answer is the first option

explanation:

just googled it m8. not that hard

lol