Question 2 of 10: Batch Fermentation Mass Balance — Ethanol and Propenoic Acid Yield

Question 2: Batch Fermentation Mass Balance — Ethanol and Propenoic Acid Yield (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Find. Weight percent of ethanol and of propenoic acid remaining in the final broth.

Approach. CO₂ is the only species produced exclusively by the ethanol-forming reaction, so its measured mass fixes how much glucose went down that path; the glucose balance then gives the remainder that went down the acid-forming path. Both product masses follow from reaction stoichiometry, and the final broth mass is the feed mass less the CO₂ that vents off as gas.

1. Feed composition. Glucose charged $=0.12(4000)=480$ kg; water charged $=3520$ kg.
2. Glucose reacted. $480-90=390$ kg $=390/180=2.1667$ kmol.
3. Split by reaction via the CO₂ yield. Only Reaction 1 produces CO₂ (2 mol CO₂ per mol glucose): $n_{\text{CO}_2}=120/44=2.7273$ kmol $\Rightarrow x_1=n_{\text{CO}_2}/2=1.3636$ kmol glucose via Reaction 1. The remainder went via Reaction 2: $x_2=2.1667-1.3636=0.8030$ kmol.
4. Product masses. $$m_{\text{ethanol}}=2x_1(46)=2(1.3636)(46)=125.45\ \text{kg},\qquad m_{\text{acid}}=2x_2(72)=2(0.8030)(72)=115.64\ \text{kg}.$$ Reaction 2 also releases water: $2x_2(18)=28.91$ kg, added to the broth's water.
5. Final broth mass. CO₂ is the only species that leaves the tank (as gas); every other product/reactant stays in the liquid broth: $$m_{\text{broth}}=4000-120=3880\ \text{kg}.$$ (Check: $90+125.45+115.64+(3520+28.91)+120=4000.0$ kg — the full mass balance closes.)
6. Weight percents. $$\%\text{ethanol}=\frac{125.45}{3880}\times100=\boxed{3.23\%},\qquad \%\text{acid}=\frac{115.64}{3880}\times100=\boxed{2.98\%}.$$