Question

C6H12O6 = 2C2H5OH + 2CO2. A 1.00-mole sample of C6H12O6 was placed in a vat with 100g of yeasr. If 62.4 grams of C2H5OH was obtained, what was the percent yield of C2H5OH

Answer 1

The answer to your question is 67.8 %

Step-by-step explanation:

Data

62.4 grams of C₂H₅OH

Percent yield

1.- Calculate the theoretical production of Ethanol

1 mol of C₆H₁₂O₆ ------------------ 2 moles of C₂H₅OH

1 mol of C₆H₁₂O₆ ------------------ x

x = (1 x 2) / 1

x = 2 moles of C₂H₅OH

2.- Convert moles to grams

Molecular mass of C₂H₅OH = (12 x 2) + (1 x 6) + (1 x 16)

= 24 + 6 + 16

= 46 g

46 g of C₂H₅OH ---------------- 1 mol

x ---------------- 2 moles

x = (2 x 46) / 1

x = 92 g of C₂H₅OH

3.- Calculate the percent yield

Percent yield =
`(62.4)/(92) x 100`

Percent yield = 67.8 %