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

C12H22O11 (aq) + H2O (g) 4 C2H5OH (l) + 4 CO2 (g)

Determine the theoretical yield and the percent yields of ethanol if 680. g sucrose undergoes fermentation and 326.5 g ethanol is obtained. theoretical _ g _ percent %

Ans: Theoretical yield = 365.8 g

% yield = 89.3%

Given reaction:

C12H22O11 + H2O → 4C2H5OH + 4CO2

Mass of sucrose = 680 g

Molar mass of sucrose = 342 g/mol

# moles of sucrose = 680/342 = 1.988 moles

Based on stoichiometry:

1 mole of sucrose forms 4 moles of ethanol

1.988 moles of sucrose will yield: 1.988*4 = 7.952 moles of ethanol

Molar mass of ethanol = 46 g/mol

Theoretical yield of ethanol = 7.952*46 = 365.8 g

% yield = (experimental/theoretical) * 100 = (326.5/365.8) * 100 = 89.3%