Subtopic: Bio-fuels and their use in farm mechanization
Question Statement:
In the biochemical fermentation of glucose to bioethanol according to Gay-Lussac's equation: $\text{C}_6\text{H}_{12}\text{O}_6 \to 2\,\text{C}_2\text{H}_5\text{OH} + 2\,\text{CO}_2$. Taking molecular weights: $\text{C} = 12$, $\text{H} = 1$, $\text{O} = 16$. The theoretical maximum mass yield of ethanol from $1000\text{ kg}$ of pure glucose in $\text{kg}$ is ________ (round off to 1 decimal place).
Answer Key:
511.1
Step-by-Step Solution & Explanation
```
Stoichiometry of Bioethanol Fermentation:
C6H12O6 ---> 2 C2H5OH + 2 CO2
```
**Step-by-step Calculation:**
1. Molecular weight of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$):
$M_{\text{glucose}} = (6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180\text{ g/mol}$
2. Molecular weight of ethanol ($\text{C}_2\text{H}_5\text{OH}$):
$M_{\text{ethanol}} = (2 \times 12) + (6 \times 1) + (1 \times 16) = 24 + 6 + 16 = 46\text{ g/mol}$
3. Mass of ethanol produced per mole of glucose (2 moles of ethanol):
$M_{\text{product}} = 2 \times 46 = 92\text{ g}$
4. Theoretical mass yield fraction:
$Y = \frac{92}{180} = 0.51111 = 51.111\%$
5. For $1000\text{ kg}$ of glucose:
$\text{Mass of ethanol} = 1000 \times 0.51111 = 511.11\text{ kg} \approx 511.1\text{ kg}$