Question

Ethanol, C₂H₅OH, is produced by the fermentation of glucose, C6H12O6, in the reaction:

C₆H₁₂O₆(aq) ==2C₂H₅OH(aq) + 2CO₂(g)

what volume in liters of CO₂ gas at 25^oC and 788mm pressure can be produced from 125mL of a 0.15 M solution of glucose? If 0.76 L of CO₂ is produced, what is the percent yield?

Answer 1

moles of glucose = molarity* Volume in L= 0.15*125/1000=0.01875

From the stoichiometry of the reaction, moles of CO2 produced= 2*0.01875=0.0375 moles

from the gas law equation, Volume of gas V= nRT/P

P= 788/760 atm (1atm =760mm Hg)=1.04 atm T= 25 deg.c= 25+273.15= 298.15K

n=0.0375 R= 0.08206

V, Volume of gas produced = 0.0375*0.08206*298.15/1.04=0.88 L

Actual volume of gas produced= 0.76L

percent yield= 100*(Actual volume of gas/ theoretical volume)=100*0.76/0.88=86.36%