if 275.0 g of glucose yields 142.9 mL of ethanol, what is the percent yield for the reaction?

Answer 1

% yield = actual yield / theoretical yield (x100%)

We are given the actual yield = 142.9 mls

We need to know the theoretical yield of ethanol from 275.0 g of glucose.

The reaction of glucose (C₆H₁₂O₆) forming ethanol (C₂H₅OH) is

C₆H₁₂O₆(l) ==> 2 C₂H₅OH(l) + 2 CO₂(g) ... fermentation of glucose

Theoretical yield of C₂H₅OH:

molar mass C₆H₁₂O₆ = 180 g / mol

molar mass C₂H₅OH = 46.1 g / mol

275.0 g C₆H₁₂O₆ x 1 mol / 180 g = 1.528 mols C₆H₁₂O₆

1.528 mol C₆H₁₂O₆ x 2 mol C₂H₅OH / mol C₆H₁₂O₆ = 3.0556 mol C₂H₅OH theoretical yield

Given a density of 0.789 g / ml for ethanol (Wikipedia), we can find the mls of C₂H₅OH

3.0556 mol C₂H₅OH x 46.1 g / mol x 1 ml / 0.789 g = 178.5 ml C₂H₅OH theoretical yield

Percent yield = 142.9 ml / 178.5 ml (x100%) = 80.06%