Question

A 1.00-mole sample of C6H12O6 was placed in a vat with 100 g of yeast. If 32.3 grams of C2H5OH was obtained, what was the percent yield of C2H5OH?

Answer 1

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:

$n = \frac{2 moles C_{2}H_{5}OH}{1 mol C_{6}H_{12}O_{6}}*1 mol C_{6}H_{12}O_{6} = 2 moles C_{2}H_{5}OH$

Now, we need to find the mass of C₂H₅OH:

$m = n*M = 2 moles*46.07 g/mol = 92.14 g$  

Finally, the percent yield of C₂H₅OH is:

$\% = \frac{32.3 g}{92.14 g}*100 = 35.06 \%$

Therefore, the percent yield of C₂H₅OH is 35.06 %.

I hope it helps you!