"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor")
n-factor is very important in redox as well as non-redox reactions.With the help of n-factor we can predict the molar ratio of the reactant species specis taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants.
In general n-factor of acid/base is number of moles of `H^(+)//OH^(-)` furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant.
Example 1:
(1)In acidic medium : `KMnO_(4) (n=5)to Mn^(2+)`
(2) In neutral medium : `KMnO_(4)(n=3) to Mn^(2+)`
(3) In basic medium : `KMnO_(4)(n=1) to Mn^(6+)`
Example 2 : `FeC_(2)O_(4)to Fe^(3+)+2CO_(2)`
Total number of moles `e^(-)` lost by 1 mole of `FeC_(2)O_(4)`
`=1+1xx2 implies 3`
In the reaction, `xVO+yFe_(2)O_(3) to FeO+V_(2)O_(5)` what is the value of x and y respectively?

Options

A. 1,1
B. 2,3
C. 3,2
D. None of these

Answer

To solve the problem of finding the values of \( x \) and \( y \) in the reaction \( x \text{VO} + y \text{Fe}_2\text{O}_3 \rightarrow \text{FeO} + \text{V}_2\text{O}_5 \), we need to balance the reaction by determining the oxidation states and the n-factors of the reactants and products.

### Step-by-Step Solution:

**Step 1: Determine the oxidation states of the elements in the reactants and products.**

- For \( \text{VO} \):
- Vanadium (V) in \( \text{VO} \) has an oxidation state of +2.

- For \( \text{Fe}_2\text{O}_3 \):
- Iron (Fe) has an oxidation state of +3.

- For \( \text{FeO} \):
- Iron (Fe) has an oxidation state of +2.

- For \( \text{V}_2\text{O}_5 \):
- Vanadium (V) has an oxidation state of +5.

**Step 2: Calculate the n-factor for each reactant.**

- For Vanadium in \( \text{VO} \):
- Change in oxidation state from +2 to +5 (increase of 3).
- Since there is 1 atom of Vanadium, the n-factor is \( 3 \times 1 = 3 \).

- For Iron in \( \text{Fe}_2\text{O}_3 \):
- Change in oxidation state from +3 to +2 (decrease of 1).
- Since there are 2 atoms of Iron, the n-factor is \( 1 \times 2 = 2 \).

**Step 3: Set up the equation based on the n-factors.**

The total change in oxidation states must be equal on both sides of the reaction. Therefore, we can set up the equation based on the n-factors:

\[ x \cdot 3 = y \cdot 2 \]

**Step 4: Balance the number of atoms for each element.**

- For Vanadium:
- On the right side, \( \text{V}_2\text{O}_5 \) has 2 Vanadium atoms.
- Therefore, \( x = 2 \).

- For Iron:
- On the right side, \( \text{FeO} \) has 1 Iron atom, and since we have \( y \) moles of \( \text{Fe}_2\text{O}_3 \), we have \( 2y \) Iron atoms.
- Thus, we need \( 2y = 2 \) (to balance with 2 from \( \text{FeO} \)).
- Therefore, \( y = 1 \).

**Step 5: Substitute the values back into the equation.**

Now we can substitute \( x \) and \( y \) back into the equation:

\[ 2 \cdot 3 = 1 \cdot 2 \]

This gives us:

\[ 6 = 2 \]

This indicates that we need to adjust our coefficients. Since we found \( x = 2 \) and \( y = 1 \), we can multiply the entire equation by 3 to balance it properly:

\[ 2 \text{VO} + 3 \text{Fe}_2\text{O}_3 \rightarrow 3 \text{FeO} + 2 \text{V}_2\text{O}_5 \]

Thus, we find:

\[ x = 2, \quad y = 3 \]

### Final Answer:
The values of \( x \) and \( y \) are \( 2 \) and \( 3 \) respectively.