Hydrazine reacts with `KIO_(3)` in presence of `HCl` as `:`
`N_(2)H_(4)+IO_(3)^(-)+2H^(+)+Cl^(-) rarrICI+N_(2)+3H_(2)O`
The equivalent masses of `N_(2)H_(4)` and `KIO_(3)` respectively are `:`

Options

A. 8 and 53.5
B. 16 and 53.5
C. 8 and 35.6
D. 8 and 87

Answer

To find the equivalent masses of hydrazine (N₂H₄) and potassium iodate (KIO₃), we will follow these steps:

### Step 1: Determine the oxidation states
1. **Hydrazine (N₂H₄)**:
- The oxidation state of hydrogen is +1.
- Let the oxidation state of nitrogen be \( x \).
- The equation for hydrazine is:
\[ 2x + 4(+1) = 0 \implies 2x + 4 = 0 \implies 2x = -4 \implies x = -2 \]
- Therefore, the oxidation state of nitrogen in N₂H₄ is -2.

2. **Potassium Iodate (KIO₃)**:
- The oxidation state of potassium (K) is +1 and oxygen (O) is -2.
- Let the oxidation state of iodine be \( y \).
- The equation for potassium iodate is:
\[ +1 + y + 3(-2) = 0 \implies +1 + y - 6 = 0 \implies y - 5 = 0 \implies y = +5 \]
- Therefore, the oxidation state of iodine in KIO₃ is +5.

### Step 2: Calculate the change in oxidation states
1. **For N₂H₄**:
- The nitrogen changes from -2 to 0 (in N₂).
- The change in oxidation state = \( 0 - (-2) = 2 \).
- Since there are 2 nitrogen atoms, the total change = \( 2 \times 2 = 4 \).

2. **For KIO₃**:
- The iodine changes from +5 to +1 (in ICl).
- The change in oxidation state = \( +1 - (+5) = -4 \).
- Since there is 1 iodine atom, the total change = \( 4 \).

### Step 3: Determine the n-factor
- The n-factor for N₂H₄ = 4 (from the change in oxidation states).
- The n-factor for KIO₃ = 4 (from the change in oxidation states).

### Step 4: Calculate the molar masses
1. **Molar mass of N₂H₄**:
- Nitrogen (N): 14 g/mol, so for 2 nitrogen atoms = \( 2 \times 14 = 28 \) g/mol.
- Hydrogen (H): 1 g/mol, so for 4 hydrogen atoms = \( 4 \times 1 = 4 \) g/mol.
- Total molar mass of N₂H₄ = \( 28 + 4 = 32 \) g/mol.

2. **Molar mass of KIO₃**:
- Potassium (K): 39 g/mol.
- Iodine (I): 127 g/mol.
- Oxygen (O): 16 g/mol, so for 3 oxygen atoms = \( 3 \times 16 = 48 \) g/mol.
- Total molar mass of KIO₃ = \( 39 + 127 + 48 = 214 \) g/mol.

### Step 5: Calculate the equivalent masses
1. **Equivalent mass of N₂H₄**:
\[ \text{Equivalent mass of N₂H₄} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{32}{4} = 8 \text{ g/equiv} \]

2. **Equivalent mass of KIO₃**:
\[ \text{Equivalent mass of KIO₃} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{214}{4} = 53.5 \text{ g/equiv} \]

### Final Answer:
The equivalent masses of N₂H₄ and KIO₃ are 8 g/equiv and 53.5 g/equiv, respectively.

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