To solve the problem, we need to follow these steps:
### Step 1: Calculate the moles of acetic acid
Given:
- Volume of acetic acid solution = 100 mL = 0.1 L
- Molarity of acetic acid = 0.1 M
Using the formula:
\[
\text{Moles of acetic acid} = \text{Molarity} \times \text{Volume (L)}
\]
\[
\text{Moles of acetic acid} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{moles}
\]
### Step 2: Determine the moles of NaOH required for neutralization
The neutralization reaction between acetic acid (CH₃COOH) and sodium hydroxide (NaOH) can be represented as:
\[
\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}
\]
From the reaction, we see that 1 mole of acetic acid reacts with 1 mole of NaOH. Therefore, the moles of NaOH required will also be:
\[
\text{Moles of NaOH} = 0.01 \, \text{moles}
\]
### Step 3: Electrolysis of the resulting solution
After neutralization, we have sodium acetate (CH₃COONa) in solution. During electrolysis, sodium acetate can produce ethane (C₂H₆) according to the following reaction:
\[
2 \text{CH}_3\text{COONa} \rightarrow \text{C}_2\text{H}_6 + 2 \text{NaOH} + \text{H}_2 + \text{CO}_2
\]
From the balanced equation, 2 moles of sodium acetate produce 1 mole of ethane.
### Step 4: Calculate the moles of ethane produced
Since we have 0.01 moles of acetic acid, it will produce:
\[
\text{Moles of sodium acetate} = 0.01 \, \text{moles}
\]
Using the stoichiometry from the electrolysis reaction:
\[
\text{Moles of ethane} = \frac{0.01}{2} = 0.005 \, \text{moles}
\]
### Step 5: Calculate the volume of ethane at STP
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L. Therefore, the volume of ethane produced is:
\[
\text{Volume of ethane} = \text{Moles of ethane} \times 22.4 \, \text{L/mol}
\]
\[
\text{Volume of ethane} = 0.005 \, \text{moles} \times 22.4 \, \text{L/mol} = 0.112 \, \text{L} = 112 \, \text{mL}
\]
### Final Answer
The volume of ethane obtained at STP after the complete electrolysis of the resulting solution is **112 mL**.
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