Limiting Reagent Question For Chemistry Souring ✓ Solved

Limiting Reagent Question for Chemistry Souring

Limiting Reagent Question for Chemistry Souring

Souring of wine occurs when ethanol is converted to acetic acid by oxygen following the reaction:

C2H5OH + O2 → CH3COOH + H2O.

A 1.00 L bottle of wine, labeled as 8.5% (by volume) ethanol, has a defective seal. Analysis of 1.00 mL of the wine shows that there are 0.0274 grams of acetic acid. The density of ethanol is 0.816 g/mL, and the density of water is 1.00 g/mL.

Part a: What mass of oxygen must have leaked into the bottle?

Part b: What is the percent yield for the conversion of ethanol to acetic acid if O2 is in excess?

Sample Paper For Above instruction

Introduction

The process of wine souring due to microbial or chemical oxidation involves the conversion of ethanol to acetic acid when oxygen leaks into the bottle. This reaction is significant in understanding spoilage of wine and estimating the extent of oxidation based on analytical data. In this paper, we analyze the given data to calculate the amount of oxygen that leaked into the bottle and determine the efficiency of the conversion process.

Calculating the Mass of Ethanol in the Wine

First, we determine the total volume of ethanol originally present in the 1.00L bottle based on the volume percentage.

Given that the wine is 8.5% ethanol by volume:

  • Volume of ethanol = 8.5% of 1.00 L = 0.085 L = 85 mL

Using the density of ethanol (0.816 g/mL), the mass of ethanol is:

Mass of ethanol = Volume × Density = 85 mL × 0.816 g/mL ≈ 69.36 g

To find moles of ethanol:

Molar mass of ethanol (C2H6O) ≈ 46.07 g/mol


Moles of ethanol = 69.36 g / 46.07 g/mol ≈ 1.506 mol

Determining the Moles of Acetic Acid Present

Analytical data indicates 0.0274 g of acetic acid in 1 mL of wine. To find total acetic acid in the entire 1 L (1000 mL) of wine:

Total acetic acid = 0.0274 g/mL × 1000 mL = 27.4 g

Molar mass of acetic acid (CH3COOH) ≈ 60.05 g/mol

Moles of acetic acid formed:

Moles of acetic acid = 27.4 g / 60.05 g/mol ≈ 0.456 mol

Calculating the Moles of Oxygen Needed

The reaction shows a 1:1 molar ratio of ethanol to oxygen:

C2H5OH + O2 → CH3COOH + H2O

Since 0.456 mol of acetic acid was produced, and assuming all originated from ethanol reacting with oxygen, the moles of ethanol that reacted are also 0.456 mol.

Mass of Oxygen Leaked Into the Bottle

Moles of oxygen that leaked in:

Oxygen molar ratio = 1:1 with ethanol


Moles of O2 = 0.456 mol


Mass of oxygen = moles × molar mass


Molar mass of oxygen (O2) ≈ 32.00 g/mol


Mass of oxygen = 0.456 mol × 32.00 g/mol ≈ 14.59 g

Thus, approximately 14.59 grams of oxygen leaked into the bottle.

Calculating the Percent Yield of the Conversion

The ideal amount of oxygen that could have reacted (if all ethanol reacted completely to form acetic acid) is based on the initial ethanol amount.

Initial moles of ethanol = 1.506 mol

If all ethanol reacted, the theoretical moles of acetic acid produced would be 1.506 mol (since it’s a 1:1 reaction).

Given that only 0.456 mol of acetic acid was actually formed, the percent yield is:

Percent yield = (actual moles / theoretical moles) × 100


= (0.456 mol / 1.506 mol) × 100 ≈ 30.3%

Therefore, the efficiency of the conversion process is approximately 30.3%, indicating that only about one-third of the available ethanol was oxidized to acetic acid.

Conclusion

This analysis demonstrates how analytical data combined with stoichiometric calculations can estimate the amount of oxygen ingress in a wine bottle and evaluate the efficiency of the souring process. Such evaluations are crucial for quality control and understanding spoilage mechanisms in wine storage.

References

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