In the disproportionation `3HClO_(3)rarrHClO_(4)+Cl_(2)+2O_(2)+H_(2)O` the equivalent mass of the oxidising agent is (molar mass of `HClO_(4) = 84.45`)

Options

A. 16.89
B. 32.22
C. 84.45
D. 28.15

Explanation

To find the equivalent mass of the oxidizing agent in the disproportionation reaction \(3HClO_3 \rightarrow HClO_4 + Cl_2 + 2O_2 + H_2O\), we will follow these steps:

### Step 1: Identify the Oxidation States
1. **Determine the oxidation state of chlorine in HClO3 and HClO4**:
- For \(HClO_3\):
- Hydrogen (H) = +1
- Oxygen (O) = -2 (3 O atoms contribute -6)
- Let the oxidation state of Cl be \(x\).
- The equation becomes: \(+1 + x - 6 = 0\) → \(x = +5\).
- For \(HClO_4\):
- Hydrogen (H) = +1
- Oxygen (O) = -2 (4 O atoms contribute -8)
- Let the oxidation state of Cl be \(y\).
- The equation becomes: \(+1 + y - 8 = 0\) → \(y = +7\).

### Step 2: Determine the Changes in Oxidation States
2. **Identify the changes in oxidation states**:
- In the reaction, Cl in \(HClO_3\) changes from +5 to +7 when it forms \(HClO_4\) (oxidation).
- Cl in \(HClO_3\) changes from +5 to 0 when it forms \(Cl_2\) (reduction).

### Step 3: Calculate the N Factor
3. **Calculate the N factor**:
- The N factor is defined as the total change in oxidation number per mole of the substance.
- For the oxidation to \(HClO_4\): Change = \(+7 - (+5) = +2\).
- For the reduction to \(Cl_2\): Change = \(+5 - 0 = +5\).
- Since there are 3 moles of \(HClO_3\) involved, the total change in oxidation number for the reduction is \(3 \times 5 = 15\).
- Therefore, the N factor for \(HClO_3\) is \(5\) (as it undergoes both oxidation and reduction).

### Step 4: Calculate the Equivalent Mass
4. **Calculate the equivalent mass**:
- The equivalent mass (or equivalent weight) is given by the formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{N factor}} \]
- The molar mass of \(HClO_4\) is given as \(84.45 \, g/mol\).
- Substitute the values into the formula:
\[ \text{Equivalent mass} = \frac{84.45}{5} = 16.89 \, g/equiv \]

### Final Answer
The equivalent mass of the oxidizing agent \(HClO_3\) is **16.89 g/equiv**.