Number of moles of `K_(2)Cr_(2)O_(7)` required to oxidised `12` moles of ethanol to acetic acid in acidic meidum will be

Explanation

To solve the problem of determining the number of moles of \( K_2Cr_2O_7 \) required to oxidize 12 moles of ethanol to acetic acid in an acidic medium, we can follow these steps:

### Step 1: Write the balanced chemical equation
The oxidation of ethanol (\( C_2H_5OH \)) to acetic acid (\( CH_3COOH \)) can be represented as:
\[ C_2H_5OH + H_2O \rightarrow CH_3COOH + 4H^+ + 4e^- \]
This equation shows that 1 mole of ethanol produces 1 mole of acetic acid and releases 4 electrons.

### Step 2: Determine the number of electrons transferred
From the balanced equation, we see that for every mole of ethanol oxidized, 4 moles of electrons are transferred. Therefore, for 12 moles of ethanol:
\[ \text{Electrons transferred} = 12 \, \text{moles of ethanol} \times 4 \, \text{moles of electrons/mole of ethanol} = 48 \, \text{moles of electrons} \]

### Step 3: Determine the equivalent weight of \( K_2Cr_2O_7 \)
In acidic medium, \( K_2Cr_2O_7 \) acts as an oxidizing agent and each mole of \( K_2Cr_2O_7 \) can accept 6 moles of electrons. Therefore, the number of equivalents of \( K_2Cr_2O_7 \) required to accept 48 moles of electrons is calculated as follows:
\[ \text{Equivalents of } K_2Cr_2O_7 = \frac{48 \, \text{moles of electrons}}{6 \, \text{moles of electrons/mole of } K_2Cr_2O_7} = 8 \, \text{equivalents} \]

### Step 4: Calculate the number of moles of \( K_2Cr_2O_7 \)
Since the number of equivalents is equal to the number of moles multiplied by the n-factor (which is 6 for \( K_2Cr_2O_7 \)), we can find the number of moles:
\[ \text{Moles of } K_2Cr_2O_7 = \frac{\text{Equivalents}}{n} = \frac{8 \, \text{equivalents}}{1} = 8 \, \text{moles} \]

### Final Answer
The number of moles of \( K_2Cr_2O_7 \) required to oxidize 12 moles of ethanol to acetic acid in acidic medium is **8 moles**.

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