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

Answer:

The answer to your question is 67.8 %

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 = $\frac{62.4}{92} x 100$

 Percent yield = 67.8 %