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MAXImum [283]
3 years ago
13

PLEASE HELPPPP???!!!!!!

Chemistry
1 answer:
sertanlavr [38]3 years ago
5 0
Mass of molecule (g) = Mr of substance over avarogado constant
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The combustion of methane(CH4) produces carbon dioxide and water. Assume that 2.0 mol of CH4 burned in the presence of excess ai
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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 %

3 0
3 years ago
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