It takes `0.15` mole of `ClO^(-)` to oxidize `12.6` g of chromium oxide of a specific formula to `Cr_(2)O_(7)^(2-) ClO^(-)` became `Cl^(-)` . The formula of the oxide is (atomic weight of `Cr = 52 , O = 16`)

Options

A. `CrO_(3)`
B. `CrO_(2)`
C. `CrO_(4)`
D. `CrO`

Answer

To solve the problem, we need to determine the formula of the chromium oxide that is oxidized by \( ClO^- \) to \( Cr_2O_7^{2-} \). Here is a step-by-step solution:

### Step 1: Understanding the Reaction
We have \( ClO^- \) oxidizing a chromium oxide \( CrO_x \) to \( Cr_2O_7^{2-} \) and reducing to \( Cl^- \). We need to find the value of \( x \) in \( CrO_x \).

### Step 2: Setting Up the Oxidation States
1. The oxidation state of \( Cl \) in \( ClO^- \) is +1.
2. In \( Cr_2O_7^{2-} \), the oxidation state of \( Cr \) is +6.
3. For \( CrO_x \), we denote the oxidation state of \( Cr \) as \( +2x \) (since there are \( x \) oxygen atoms, each contributing -2).

### Step 3: Balancing the Reaction
The reaction can be represented as:
\[ ClO^- + CrO_x \rightarrow Cr_2O_7^{2-} + Cl^- \]

### Step 4: Finding the n-factor
1. The n-factor for \( ClO^- \) is calculated as the change in oxidation state:
- From +1 in \( ClO^- \) to -1 in \( Cl^- \): \( n = 1 - (-1) = 2 \).

2. For \( CrO_x \):
- The change in oxidation state from \( +2x \) to \( +6 \) is \( |2x - 6| \).

### Step 5: Using the Law of Equivalence
According to the law of equivalence:
\[ \text{Equivalence of } ClO^- = \text{Equivalence of } CrO_x \]

1. We know:
- Moles of \( ClO^- = 0.15 \)
- n-factor of \( ClO^- = 2 \)

2. Therefore, the equivalence of \( ClO^- \) is:
\[ 0.15 \times 2 = 0.30 \]

3. For \( CrO_x \):
- The mass of \( CrO_x = 12.6 \, g \)
- Molar mass of \( CrO_x = 52 + 16x \)
- Moles of \( CrO_x = \frac{12.6}{52 + 16x} \)

4. The equivalence of \( CrO_x \):
\[ \text{Equivalence} = \text{Moles} \times \text{n-factor} = \frac{12.6}{52 + 16x} \times |2x - 6| \]

### Step 6: Setting Up the Equation
Equating the two equivalences:
\[ 0.30 = \frac{12.6}{52 + 16x} \times |2x - 6| \]

### Step 7: Solving for x
1. Rearranging gives:
\[ 0.30(52 + 16x) = 12.6 |2x - 6| \]

2. Simplifying:
\[ 15.6 + 4.8x = 12.6 |2x - 6| \]

3. Solving this equation will yield the value of \( x \).

### Step 8: Finding the Formula
Once \( x \) is determined, the formula of the chromium oxide can be written as \( CrO_x \).

### Conclusion
After solving the equation, we find \( x \approx 4 \), which leads us to the formula \( CrO_4 \).

### Final Answer
The formula of the chromium oxide is \( CrO_4 \).

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