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Consider the fermentation reaction of glucose: C6H12O6 → 2C2H5OH + 2CO2 A 1.00-mol sample of C6H12O6 was placed in a vat with 100 g of yeast. If 67.7 g of C2H5OH was obtained, what was the percent yield of C2H5OH?

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whitneytr12

Answer:

% yield = 73.48 %

Explanation:

The fermentation reaction is:

C₆H₁₂O₆  →  2C₂H₅OH + 2CO₂          

The percent yield of C₂H₅OH is given by:

[tex] \% yield = \frac{m_{E}}{m_{T}} * 100 [/tex]

where [tex]m_{E}[/tex]: is the obtained mass of C₂H₅OH = 67.7g and [tex]m_{T}[/tex]: is the theoretical mass of C₂H₅OH.    

The theoretical mass of C₂H₅OH is calculated knowing that 1 mol of C₆H₁₂O₆ produces 2 moles of C₂H₅OH:  

[tex] m_{T} = mol * M [/tex]

where M: is the molar mass of C₂H₅OH =  46.068 g/mol

[tex] m_{T} = 2 moles * 46.068 g/mol = 92.136 g [/tex]                

Hence, the percent yield of C₂H₅OH is:

[tex] \% yield = \frac{67.7 g}{92.136 g}*100 = 73.48 \% [/tex]

I hope it helps you!                  

CintiaSalazar

Taking into account definition of percent yield, the percent yield for the reaction is 73.58%.

Reaction stoichiometry

In first place, the balanced reaction is:

C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • C₆H₁₂O₆: 1 mole
  • C₂H₅OH: 2 moles
  • CO₂: 2 moles

The molar mass of the compounds is:

  • C₆H₁₂O₆: 180 g/mole
  • C₂H₅OH: 46 g/mole
  • CO₂: 44 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • C₆H₁₂O₆: 1 moles× 180 g/mole= 180 grams
  • C₂H₅OH: 2 moles× 46 g/mole= 92 grams
  • CO₂: 2 moles× 44 g/mole= 88 grams

Percent yield

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

[tex]percent yield=\frac{actual yield}{theoretical yield} x100[/tex]

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

Percent yield for the reaction in this case

In this case, you know:

  • actual yield= 67.7 grams
  • theorical yield= 92 grams

Replacing in the definition of percent yield:

[tex]percent yield=\frac{67.7 g}{92 g} x100[/tex]

Solving:

percent yield= 73.58%

Finally, the percent yield for the reaction is 73.58%.

Learn more about

the reaction stoichiometry:

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percent yield:

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