The heat of neutralisation of strong base and strong acid is 57.0kJ/mol. The heat released when 0.5 mole of `HNO_(3)` is added to 0.20 mole of NaOH solution is

Options

A. 34.9kJ
B. 57.0kJ
C. 11.4kJ
D. 28.5kJ

Explanation

To solve the problem, we need to determine the heat released when 0.5 moles of HNO3 is added to 0.2 moles of NaOH, 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 reaction between a strong acid (HNO3) and a strong base (NaOH) can be represented as:
\[ \text{HNO}_3 + \text{NaOH} \rightarrow \text{NaNO}_3 + \text{H}_2\text{O} \]
This reaction produces salt (sodium nitrate) and water.

2. **Determine the Moles of Reactants**:
- Moles of HNO3 = 0.5 moles
- Moles of NaOH = 0.2 moles

3. **Identify the Limiting Reactant**:
The limiting reactant is the one that will be completely consumed first in the reaction. In this case:
- HNO3 can react with NaOH in a 1:1 ratio.
- Since we have 0.2 moles of NaOH and 0.5 moles of HNO3, NaOH is the limiting reactant.

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

5. **Final Answer**:
The heat released when 0.5 moles of HNO3 is added to 0.2 moles of NaOH is **11.4 kJ**.