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A balanced chemical equation shows the relationship between reactants and products in a chemical reaction. It reflects the conservation of mass, meaning the number of atoms per element in reactants must equal the number in the products. In the fermentation reaction, glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) is converted into ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) and carbon dioxide (\(\text{CO}_2\)).
The unbalanced form of this reaction might look like: - \[\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{CO}_2\] By balancing, we adjust coefficients to ensure equality of each element on both sides. Therefore, the balanced equation becomes:

• \[\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2\]

This shows that one glucose molecule produces two molecules of ethanol and two of carbon dioxide. Having a balanced equation is crucial for proceeding further with yield calculations.

Theoretical yield is the maximum amount of product expected from a reaction. It is calculated using the balanced equation and the reactant amounts. By using molecular weights and stoichiometry, you'll determine the potential output, assuming perfect conditions.
**Step-by-Step Process**

• First, calculate the molar mass of glucose by summing the atomic masses of its constituents: - Carbon: 6 × 12.01 g/mol = 72.06 g/mol- Hydrogen: 12 × 1.01 g/mol = 12.12 g/mol- Oxygen: 6 × 16.00 g/mol = 96.00 g/mol- Total for glucose = 180.18 g/mol
• For ethanol, sum similarly: - Carbon: 2 × 12.01 g/mol = 24.02 g/mol- Hydrogen: 6 × 1.01 g/mol = 6.06 g/mol- Oxygen: 1 × 16.00 g/mol = 16.00 g/mol- Total for ethanol = 46.08 g/mol
• Convert 100 g of glucose to moles: \[100 \text{ g} \times \left( \frac{1 \text{ mol}}{180.18 \text{ g}} \right) = 0.555 \text{ moles} \]
• Using the balanced equation (1 mole glucose = 2 moles ethanol), these 0.555 moles yield \(0.555 \times 2 = 1.11\) moles of ethanol.
  
• Finally, convert to grams: \[1.11 \text{ moles} \times 46.08 \text{ g/mol} = 51.13 \text{ g} \]

This 51.13 g is the theoretical yield of ethanol.

Percent yield assesses the efficiency of a chemical reaction, comparing the actual yield to the theoretical yield. It indicates how much product was actually obtained from the reaction relative to the maximum possible amount. This is a valuable measure in experiments.
**Steps to Calculate Percent Yield:**

• Determine the actual yield of ethanol. We're given 50.0 mL of ethanol, which is then converted to mass using the density provided (0.789 g/mL): \[\text{Mass} = 50.0 \text{ mL} \times 0.789 \text{ g/mL} = 39.45 \text{ g}\]
• Compare this to the theoretical yield determined earlier: 51.13 g.
  
• Apply the percent yield formula: \[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\]
- Substituting the values: \[\text{Percent Yield} = \left( \frac{39.45 \text{ g}}{51.13 \text{ g}} \right) \times 100 \approx 77.13\%\]

This helps in identifying losses or inefficiencies during the process and improving future reactions.