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 acidic medium, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Reaction**:
The oxidation of ethanol (\( C_2H_5OH \)) to acetic acid (\( CH_3COOH \)) can be represented as:
\[ C_2H_5OH + H_2O \xrightarrow{K_2Cr_2O_7} CH_3COOH + 4H^+ + 4e^- \]
This indicates that 1 mole of ethanol produces 1 mole of acetic acid and releases 4 electrons.

2. **Determine the Electrons Transferred**:
Since 12 moles of ethanol are being oxidized, the total number of electrons transferred in the reaction will be:
\[ \text{Total electrons} = 12 \text{ moles of ethanol} \times 4 \text{ electrons/mole} = 48 \text{ electrons} \]

3. **Find the Equivalent of \( K_2Cr_2O_7 \)**:
The balanced reaction for \( K_2Cr_2O_7 \) in acidic medium shows that it can accept 6 electrons per mole. Therefore, the number of moles of \( K_2Cr_2O_7 \) needed to accept 48 electrons can be calculated as follows:
\[ \text{Moles of } K_2Cr_2O_7 = \frac{\text{Total electrons}}{\text{Electrons accepted per mole}} = \frac{48}{6} = 8 \text{ moles} \]

### 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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