x mmol of `KMnO_(4)` react completely with y mmol of `MnSO_(4)` in presence of fluoride ions to give `MnF_(4)` quantitatively. Then :

Options

A. x=y
B. 4x=y
C. `xgty`
D. `x lt y`

Answer

To solve the problem, we need to analyze the stoichiometry of the reaction between potassium permanganate (KMnO₄) and manganese(II) sulfate (MnSO₄) in the presence of fluoride ions (F⁻) to produce manganese(IV) fluoride (MnF₄).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction involves KMnO₄ and MnSO₄ in the presence of fluoride ions. The balanced reaction can be represented as:
\[ \text{KMnO}_4 + \text{MnSO}_4 + \text{F}^- \rightarrow \text{MnF}_4 + \text{other products} \]

2. **Determine the Stoichiometry**:
From the information provided, it is known that 3 moles of KMnO₄ react with 2 moles of MnSO₄. Thus, we can write the stoichiometric relationship:
\[ 3 \text{ moles of KMnO}_4 \text{ react with } 2 \text{ moles of MnSO}_4 \]

3. **Set Up the Equivalence**:
Let \( x \) be the millimoles of KMnO₄ and \( y \) be the millimoles of MnSO₄. According to the stoichiometry:
\[ \frac{x}{3} = \frac{y}{2} \]
Cross-multiplying gives:
\[ 2x = 3y \]

4. **Rearranging the Equation**:
Rearranging the equation \( 2x = 3y \) to express \( y \) in terms of \( x \):
\[ y = \frac{2}{3}x \]

5. **Analyzing the Relationship**:
From the equation \( y = \frac{2}{3}x \), we can see that for any positive value of \( x \), \( y \) will always be less than \( x \). This implies:
\[ y < x \]

6. **Conclusion**:
Therefore, the final conclusion is that \( y \) is less than \( x \).

### Final Answer:
- \( y < x \)