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

Explanation

To solve the problem, we need to calculate the heat released when 0.5 moles of nitric acid (HNO₃) is added to 0.20 moles of sodium hydroxide (NaOH) solution, given that the heat of neutralization for a strong acid and strong base is 57.0 kJ per mole.

### Step-by-Step Solution:

1. **Identify the limiting reagent**:
- We have 0.5 moles of HNO₃ and 0.20 moles of NaOH.
- The neutralization reaction is:
\[ \text{HNO}_3 + \text{NaOH} \rightarrow \text{NaNO}_3 + \text{H}_2\text{O} \]
- Since the reaction is a 1:1 ratio, we need to determine which reactant will be used up first (the limiting reagent).
- Here, NaOH is the limiting reagent because we have only 0.20 moles of it, while we have excess HNO₃ (0.5 moles).

2. **Calculate the amount of heat released**:
- The heat of neutralization is given as 57.0 kJ per mole.
- Since NaOH is the limiting reagent, we will use the amount of NaOH to calculate the heat released.
- The amount of heat released for the neutralization of 0.20 moles of NaOH is:
\[ \text{Heat released} = \text{moles of NaOH} \times \text{heat of neutralization} \]
\[ \text{Heat released} = 0.20 \, \text{moles} \times 57.0 \, \text{kJ/mole} = 11.4 \, \text{kJ} \]

3. **Final Result**:
- The total heat released when 0.5 moles of HNO₃ is added to 0.20 moles of NaOH is **11.4 kJ**.