the number of moles of `Cr_(2)O_(7)^(2)` needed to oxidise `0.136` equivalent of `N_(2)H_(5)` through the reaction:
`N_(2)H_(5) + Cr_(2)O_(7)^(2-) rarr N_(2) + Cr^(3) + H_(2)O` is:

Options

A. `0.0226`
B. `0.087`
C. `0.136`
D. `0.488`

Explanation

To solve the problem of determining the number of moles of `Cr_(2)O_(7)^(2-)` needed to oxidize `0.136` equivalent of `N_(2)H_(5)`, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: The reaction given is:
\[ N_{2}H_{5} + Cr_{2}O_{7}^{2-} \rightarrow N_{2} + Cr^{3+} + H_{2}O \]
Here, `N_(2)H_(5)` is being oxidized by `Cr_(2)O_(7)^(2-)`.

2. **Equivalence Concept**: In this reaction, the equivalence of `N_(2)H_(5)` is equal to the equivalence of `Cr_(2)O_(7)^(2-)`. Given that the equivalence of `N_(2)H_(5)` is `0.136`, we can conclude:
\[ \text{Equivalence of } Cr_{2}O_{7}^{2-} = 0.136 \]

3. **N Factor Calculation**: The n-factor for `Cr_(2)O_(7)^(2-)` is determined by the change in oxidation state of chromium. In `Cr_(2)O_(7)^(2-)`, the oxidation state of chromium is +6. When it is reduced to `Cr^(3+)`, the change in oxidation state is:
\[ \text{Change} = 6 - 3 = 3 \]
Since there are 2 chromium atoms in `Cr_(2)O_(7)^(2-)`, the total n-factor is:
\[ \text{n-factor} = 2 \times 3 = 6 \]

4. **Relate Equivalence to Moles**: The relationship between equivalence, number of moles, and n-factor is given by:
\[ \text{Equivalence} = \text{Number of moles} \times \text{n-factor} \]
Rearranging this gives us:
\[ \text{Number of moles} = \frac{\text{Equivalence}}{\text{n-factor}} \]

5. **Calculate Number of Moles**: Now we can substitute the values we have:
\[ \text{Number of moles of } Cr_{2}O_{7}^{2-} = \frac{0.136}{6} \]
Performing the calculation:
\[ \text{Number of moles of } Cr_{2}O_{7}^{2-} = 0.0226667 \approx 0.0227 \]

6. **Final Answer**: Therefore, the number of moles of `Cr_(2)O_(7)^(2-)` needed to oxidize `0.136` equivalent of `N_(2)H_(5)` is approximately:
\[ 0.0227 \text{ moles} \]