The ratio of number of moles of `KMnO_(4)" and "K_(2)Cr_(2)O_(7)` required to oxidise 0.1 mol `Sn^(2+)" to " Sn^(+4)` in acidic medium

Options

A. `6:5`
B. `5:6`
C. `1:2`
D. `2:1`

Answer

To solve the problem of finding the ratio of the number of moles of \( KMnO_4 \) and \( K_2Cr_2O_7 \) required to oxidize \( 0.1 \) mol of \( Sn^{2+} \) to \( Sn^{4+} \) in acidic medium, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Change in Oxidation State for \( Sn^{2+} \):**
- The oxidation state of \( Sn^{2+} \) is changing to \( Sn^{4+} \).
- The change in oxidation state is \( 4 - 2 = 2 \).
- Therefore, the valence factor (n) for \( Sn^{2+} \) is \( 2 \).

2. **Determine the Moles of \( KMnO_4 \) Required:**
- In acidic medium, \( KMnO_4 \) is reduced from \( MnO_4^- \) to \( Mn^{2+} \).
- The change in oxidation state for \( Mn \) is \( 7 \) to \( 2 \), which is a change of \( 5 \).
- Therefore, the valence factor for \( KMnO_4 \) is \( 5 \).
- Using the formula for equivalents:
\[ \text{Equivalents of } Sn^{2+} = \text{Equivalents of } KMnO_4 \]
\[ n \times \text{moles of } Sn^{2+} = n \times \text{moles of } KMnO_4 \]
\[ 2 \times 0.1 = 5 \times \text{moles of } KMnO_4 \]
\[ \text{moles of } KMnO_4 = \frac{0.2}{5} = 0.04 \text{ moles} \]

3. **Determine the Moles of \( K_2Cr_2O_7 \) Required:**
- In acidic medium, \( K_2Cr_2O_7 \) is reduced from \( Cr_2O_7^{2-} \) to \( Cr^{3+} \).
- The change in oxidation state for \( Cr \) is \( 6 \) to \( 3 \), which is a change of \( 3 \).
- Therefore, the valence factor for \( K_2Cr_2O_7 \) is \( 6 \).
- Using the formula for equivalents:
\[ \text{Equivalents of } Sn^{2+} = \text{Equivalents of } K_2Cr_2O_7 \]
\[ 2 \times 0.1 = 6 \times \text{moles of } K_2Cr_2O_7 \]
\[ \text{moles of } K_2Cr_2O_7 = \frac{0.2}{6} \approx 0.0333 \text{ moles} \]

4. **Calculate the Ratio of Moles of \( KMnO_4 \) to \( K_2Cr_2O_7 \):**
- The ratio of moles of \( KMnO_4 \) to \( K_2Cr_2O_7 \) is given by:
\[ \text{Ratio} = \frac{\text{moles of } KMnO_4}{\text{moles of } K_2Cr_2O_7} = \frac{0.04}{0.0333} \]
- Simplifying this ratio:
\[ \text{Ratio} = \frac{0.04 \times 6}{0.2} = \frac{0.24}{0.2} = 1.2 \]
- This can be expressed as \( \frac{6}{5} \).

### Final Answer:
The ratio of the number of moles of \( KMnO_4 \) to \( K_2Cr_2O_7 \) required to oxidize \( 0.1 \) mol of \( Sn^{2+} \) to \( Sn^{4+} \) in acidic medium is \( \frac{6}{5} \).