What volume of`3MHNO_(3)` is needed to oxidize 8 g`Fe(2+)toFe(3+),HNO_(3)` of to gets converted to NO ?

Options

A. 8 Ml
B. 16 mL
C. 32 mL
D. 64 mL

Explanation

To solve the problem of determining the volume of 3 M HNO₃ needed to oxidize 8 g of Fe²⁺ to Fe³⁺, while HNO₃ is converted to NO, we can follow these steps:

### Step 1: Determine the n-factor of HNO₃
The n-factor of a substance in a redox reaction is defined as the number of electrons gained or lost per molecule.

1. The oxidation state of nitrogen in HNO₃ is +5.
2. In NO, the oxidation state of nitrogen is +2.
3. The change in oxidation state for nitrogen is from +5 to +2, which means it gains 3 electrons.

Thus, the n-factor of HNO₃ is **3**.

### Step 2: Determine the n-factor of Fe²⁺
For the oxidation of Fe²⁺ to Fe³⁺:

1. The oxidation state of Fe changes from +2 to +3, which means it loses 1 electron.

Thus, the n-factor of Fe²⁺ is **1**.

### Step 3: Calculate the gram equivalence of Fe²⁺
The gram equivalence can be calculated using the formula:
\[ \text{Gram Equivalence} = \frac{\text{Weight}}{\text{Molar Mass}} \times \text{n-factor} \]
Given:
- Weight of Fe²⁺ = 8 g
- Molar mass of Fe = 56 g/mol
- n-factor of Fe²⁺ = 1

Calculating the gram equivalence:
\[ \text{Gram Equivalence of Fe}^{2+} = \frac{8 \text{ g}}{56 \text{ g/mol}} \times 1 = \frac{8}{56} = \frac{1}{7} \text{ equivalents} \]

### Step 4: Set up the equation for HNO₃
Using the equivalence concept, we can equate the gram equivalences of HNO₃ and Fe²⁺:
\[ \text{Gram Equivalence of HNO}_3 = \text{Gram Equivalence of Fe}^{2+} \]
The gram equivalence of HNO₃ can be calculated as:
\[ \text{Gram Equivalence of HNO}_3 = \text{Molarity} \times \text{Volume} \times \text{n-factor} \]
Given:
- Molarity of HNO₃ = 3 M
- n-factor of HNO₃ = 3

Setting the equation:
\[ 3 \times V \times 3 = \frac{1}{7} \]
Where V is the volume in liters.

### Step 5: Solve for V
\[ 9V = \frac{1}{7} \]
\[ V = \frac{1}{63} \text{ L} \]

### Step 6: Convert volume to milliliters
To convert liters to milliliters, multiply by 1000:
\[ V = \frac{1}{63} \times 1000 \approx 15.87 \text{ mL} \]

### Step 7: Round off the final answer
Rounding off 15.87 mL gives approximately **16 mL**.

### Final Answer
The volume of 3 M HNO₃ needed to oxidize 8 g of Fe²⁺ to Fe³⁺ is approximately **16 mL**.

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