Question

Ethanol (C2H5OH) is produced from the fermentation of sucrose (table sugar) C12H22O11 in the presence of enzymes.

The balanced equation for this reaction is: C12H22O11 (aq) + H2O(g) ---> 4C2H5OH(l) + 4CO2(g)

a. What is the theoretical yield of ethanol when 648g of sucrose undergo fermentation? Show mass to mass conversion setup with units and correct sigfigs in the final answer.

b. If the actual yield is 339g, what is the percent yield? Show % yield formula setup with units and the final answer in correct sigfigs.

Answer 1

The theoretical yield of ethanol from fermenting 648g of sucrose is 348.64 g. The percent yield, with an actual yield of 339g, is 97.24%.

Step-by-step explanation:

To calculate the theoretical yield of ethanol from the fermentation of sucrose, first, we need to find the molar mass of sucrose (C12H22O11) and ethanol (C2H5OH). Using the atomic masses from the periodic table, the molar mass of sucrose is approximately 342.30 g/mol, and the molar mass of ethanol is approximately 46.07 g/mol.

To find the theoretical yield:

1. Convert mass of sucrose to moles: 648 g sucrose * (1 mol / 342.30 g) = 1.893 moles sucrose.
2. Use the stoichiometry of the balanced equation: 1.893 moles sucrose * (4 moles ethanol / 1 mole sucrose) = 7.572 moles ethanol.
3. Convert moles of ethanol to grams: 7.572 moles ethanol * (46.07 g / mol) = 348.64 g ethanol.

The theoretical yield of ethanol is 348.64 g.

Now, to calculate the percent yield:

• Percent yield = (Actual yield / Theoretical yield) * 100.
• Percent yield = (339 g / 348.64 g) * 100 = 97.24%.

The percent yield of ethanol is 97.24%.