Question

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?

Answer 1

Answer:

% yield = 73.48 %

Explanation:

The fermentation reaction is:

C₆H₁₂O₆  →  2C₂H₅OH + 2CO₂          

The percent yield of C₂H₅OH is given by:

$\% yield = \frac{m_{E}}{m_{T}} * 100$

where $m_{E}$: is the obtained mass of C₂H₅OH = 67.7g and $m_{T}$: is the theoretical mass of C₂H₅OH.    

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

$m_{T} = mol * M$

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

$m_{T} = 2 moles * 46.068 g/mol = 92.136 g$                

Hence, the percent yield of C₂H₅OH is:

$\% yield = \frac{67.7 g}{92.136 g}*100 = 73.48 \%$

I hope it helps you!