At a particlular temperature
`H^(o+) (aq) +OH^(Theta)(aq) rarr H_(2)O(l),DeltaH =- 57.1 kJ`
The approximate heat evolved when `400mL` of `0.2M H_(2)SO_(4)` is mixed with `600mL` of `0.1M KOH` solution will be

Options

A. `3.426 kJ`
B. `13.7kJ`
C. `5.2 kJ`
D. `55 kJ`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of H⁺ from H₂SO₄
Given:
- Volume of H₂SO₄ = 400 mL = 0.400 L
- Molarity of H₂SO₄ = 0.2 M

Using the formula for moles:
\[ \text{Moles of H₂SO₄} = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.400 \, \text{L} \times 0.2 \, \text{mol/L} = 0.08 \, \text{mol} \]

Since each molecule of H₂SO₄ provides 2 H⁺ ions:
\[ \text{Moles of H⁺} = 0.08 \, \text{mol} \times 2 = 0.16 \, \text{mol} \]

### Step 2: Calculate the moles of OH⁻ from KOH
Given:
- Volume of KOH = 600 mL = 0.600 L
- Molarity of KOH = 0.1 M

Using the formula for moles:
\[ \text{Moles of KOH} = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.600 \, \text{L} \times 0.1 \, \text{mol/L} = 0.06 \, \text{mol} \]

### Step 3: Determine the limiting reactant
We have:
- Moles of H⁺ = 0.16 mol
- Moles of OH⁻ = 0.06 mol

Since OH⁻ is the limiting reactant (less moles), it will react completely with H⁺.

### Step 4: Calculate the heat evolved
The heat evolved during the neutralization reaction is given by:
\[ \Delta H = -57.1 \, \text{kJ/mol} \]

The heat evolved for the reaction can be calculated using the moles of OH⁻ (since it is the limiting reactant):
\[ \text{Heat evolved} = \text{Moles of OH⁻} \times \Delta H = 0.06 \, \text{mol} \times (-57.1 \, \text{kJ/mol}) = -3.426 \, \text{kJ} \]

### Final Answer
The approximate heat evolved when 400 mL of 0.2 M H₂SO₄ is mixed with 600 mL of 0.1 M KOH is **3.426 kJ** (heat released).

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