The heat of neutralisation of a strong base and a strong acid is 13.7 kcal. The heat released when 0.6 mole HCl solution is added to 0.25 of NaOH is

Options

A. 3.425 kcal
B. 8.22 kcal
C. 11.645 kcal
D. 13.7 kcal

Answer

To solve the problem, we need to determine the heat released when 0.6 moles of HCl are added to 0.25 moles of NaOH, given that the heat of neutralization for a strong acid and a strong base is 13.7 kcal.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between HCl (strong acid) and NaOH (strong base) can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
This reaction produces water and salt (NaCl).

2. **Determine the Limiting Reactant**:
- We have 0.6 moles of HCl and 0.25 moles of NaOH.
- Since the reaction requires 1 mole of HCl for every 1 mole of NaOH, we need to determine which reactant will be completely consumed first.
- In this case, NaOH is the limiting reactant because we have less of it (0.25 moles).

3. **Calculate the Heat Released**:
- The heat of neutralization is given as 13.7 kcal for the reaction of 1 mole of a strong acid with 1 mole of a strong base.
- Since NaOH is the limiting reactant, only 0.25 moles of NaOH will react with an equivalent amount of HCl.
- The heat released for the reaction of 0.25 moles of NaOH can be calculated as follows:
\[ \text{Heat released} = \text{Heat of neutralization} \times \text{moles of NaOH} \]
\[ \text{Heat released} = 13.7 \, \text{kcal} \times 0.25 \, \text{moles} = 3.425 \, \text{kcal} \]

4. **Final Answer**:
The heat released when 0.6 moles of HCl are added to 0.25 moles of NaOH is **3.425 kcal**.