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 neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

2. **Determine the Limiting Reagent**:
From the problem, we have:
- 0.6 moles of HCl
- 0.25 moles of NaOH

Since the reaction requires 1 mole of HCl for every 1 mole of NaOH, we can see that NaOH is the limiting reagent because we have less of it (0.25 moles) compared to HCl (0.6 moles).

3. **Calculate the Heat Released**:
The heat of neutralization for the reaction is given as 13.7 kcal for the complete reaction of 1 mole of HCl with 1 mole of NaOH. Since NaOH is the limiting reagent, we will calculate the heat released based on the amount of NaOH available.

Using the formula:
\[ \text{Heat released} = \text{Heat of neutralization} \times \text{moles of limiting reagent} \]
Substituting the values:
\[ \text{Heat released} = 13.7 \, \text{kcal} \times 0.25 \, \text{moles} = 3.425 \, \text{kcal} \]

4. **Final Answer**:
The heat released when 0.6 moles of HCl are added to 0.25 moles of NaOH is:
\[ \boxed{3.425 \, \text{kcal}} \]