Answer:
y = 35.06 %.
Explanation:
The reaction of fermentation is:
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ (1)
From the reaction (1) we have that 1 mol of C₆H₁₂O₆ produces 2 moles of C₂H₅OH, then the number of moles of C₂H₅OH is:
[tex]n = \frac{2 moles C_{2}H_{5}OH}{1 mol C_{6}H_{12}O_{6}}*1 mol C_{6}H_{12}O_{6} = 2 moles C_{2}H_{5}OH[/tex]
Now, we need to find the mass of C₂H₅OH:
[tex]m = n*M = 2 moles*46.07 g/mol = 92.14 g[/tex]
Finally, the percent yield of C₂H₅OH is:
[tex] \% = \frac{32.3 g}{92.14 g}*100 = 35.06 \% [/tex]
Therefore, the percent yield of C₂H₅OH is 35.06 %.
I hope it helps you!
The percent yield of C2H5OH should be 35.06% also it have the 1.00 moles.
Since
The reaction of fermentation should be
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ (1)
Here 1 mol of C₆H₁₂O₆ generated 2 moles of C₂H₅OH, so the number of moles of C₂H₅OH should be
= 2 moles / 1mole * 1 mole
= 2 moles C₂H₅OH
So here the percent yield should be
= 32.3 / (2 * 4607)
= 32.3 / 92.14
= 35.06%
Hence, The percent yield of C2H5OH should be 35.06% also it have the 1.00 moles.
Learn more about moles here: https://brainly.com/question/19669595
