The heat of neutrasation of a strong acid and a strong alkali is `57.0KJ mol^(-1)` . The heat released when `0.5` mole of `HNO_(3)` solution is mixed with `0.2` mole of `KOH` is

Options

A. 11.5 kJ
B. 34.5 kJ
C. 23.5 kJ
D. 58.8 kJ

Explanation

To solve the problem, we need to calculate the heat released when 0.5 moles of HNO₃ is mixed with 0.2 moles of KOH, given that the heat of neutralization for a strong acid and a strong base is 57.0 kJ/mol.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The neutralization reaction between nitric acid (HNO₃) and potassium hydroxide (KOH) can be represented as:
\[ \text{HNO}_3 + \text{KOH} \rightarrow \text{KNO}_3 + \text{H}_2\text{O} \]

2. **Determine the Limiting Reagent**:
- We have 0.5 moles of HNO₃ and 0.2 moles of KOH.
- In this reaction, KOH is the limiting reagent because it will be completely consumed first.

3. **Calculate the Moles of Acid Neutralized**:
- Since KOH is the limiting reagent, it will neutralize an equal amount of HNO₃. Therefore, 0.2 moles of KOH will neutralize 0.2 moles of HNO₃.
- This leaves us with:
\[ \text{Excess HNO}_3 = 0.5 \text{ moles} - 0.2 \text{ moles} = 0.3 \text{ moles} \]

4. **Calculate the Heat Released**:
- The heat of neutralization is given as 57.0 kJ/mol.
- Since only 0.2 moles of HNO₃ are neutralized, we can calculate the heat released as follows:
\[ \text{Heat released} = \text{moles of KOH} \times \text{heat of neutralization} \]
\[ \text{Heat released} = 0.2 \text{ moles} \times 57.0 \text{ kJ/mol} = 11.4 \text{ kJ} \]

5. **Final Answer**:
The heat released when 0.5 moles of HNO₃ is mixed with 0.2 moles of KOH is **11.4 kJ**.