When `50 cm^(3)` of `0.2 N H_(2)SO_(4)` is mixed with `50 cm^(3)` of `1 N KOH`, the heat liberated is

Options

A. `11.46 kJ`
B. `57.3 kJ`
C. `573 kJ`
D. `573J`

Explanation

To solve the problem of calculating the heat liberated when 50 cm³ of 0.2 N H₂SO₄ is mixed with 50 cm³ of 1 N KOH, we can follow these steps:

### Step 1: Identify the Limiting Reagent
We have two solutions:
- 50 cm³ of 0.2 N H₂SO₄
- 50 cm³ of 1 N KOH

Since the normality of KOH is higher than that of H₂SO₄, H₂SO₄ will be the limiting reagent.

### Step 2: Calculate the Amount of H₂SO₄
The amount of H₂SO₄ in equivalents can be calculated using the formula:
\[ \text{Equivalents of H₂SO₄} = \text{Normality} \times \text{Volume (L)} \]
\[ = 0.2 \, \text{N} \times 0.050 \, \text{L} = 0.01 \, \text{equivalents} \]

### Step 3: Calculate the Amount of KOH Required
Using the stoichiometry of the neutralization reaction:
\[ \text{H₂SO₄} + 2 \, \text{KOH} \rightarrow \text{K₂SO₄} + 2 \, \text{H₂O} \]
1 equivalent of H₂SO₄ reacts with 2 equivalents of KOH. Therefore, the amount of KOH needed to neutralize 0.01 equivalents of H₂SO₄ is:
\[ \text{Equivalents of KOH} = 2 \times 0.01 = 0.02 \, \text{equivalents} \]

### Step 4: Calculate the Volume of KOH Used
Using the normality of KOH:
\[ \text{Volume of KOH (L)} = \frac{\text{Equivalents of KOH}}{\text{Normality}} = \frac{0.02}{1} = 0.02 \, \text{L} = 20 \, \text{cm}^3 \]

### Step 5: Determine the Heat Released
The heat released during the neutralization of KOH with H₂SO₄ is known to be approximately 57.3 kJ for 1 equivalent of KOH. Since we are using 0.02 equivalents of KOH:
\[ \text{Heat released (kJ)} = 57.3 \, \text{kJ} \times 0.02 = 1.146 \, \text{kJ} \]
Converting this to joules:
\[ 1.146 \, \text{kJ} = 1146 \, \text{J} \]

### Step 6: Conclusion
The heat liberated when 50 cm³ of 0.2 N H₂SO₄ is mixed with 50 cm³ of 1 N KOH is approximately **1146 J**.

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