The value of heat generated when 36.5 gm of HCI and 40 gm of NaOH reacts during neutralization

Options

A. 76.5 Kcal
B. 13.7 Kcal
C. More than 13.7 Kcal
D. 108 Kcal

Answer

To determine the heat generated when 36.5 g of HCl reacts with 40 g of NaOH during neutralization, we can follow these steps:

### Step 1: Calculate the number of moles of HCl and NaOH

1. **Molar Mass Calculation**:
- Molar mass of HCl = 1 (H) + 35.5 (Cl) = 36.5 g/mol
- Molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol

2. **Moles Calculation**:
- Moles of HCl = Mass / Molar Mass = 36.5 g / 36.5 g/mol = 1 mole
- Moles of NaOH = Mass / Molar Mass = 40 g / 40 g/mol = 1 mole

### Step 2: Identify the reaction

The neutralization reaction between HCl and NaOH can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

### Step 3: Determine the heat of neutralization

The standard heat of neutralization for strong acids and strong bases is a constant value. For the reaction between HCl and NaOH:
- The heat generated (ΔH) is approximately 13.7 kcal per mole of acid-base reaction.

### Step 4: Calculate the total heat generated

Since both HCl and NaOH are present in 1 mole each, the total heat generated during the reaction will be:
\[ \text{Total Heat} = \text{Heat per mole} \times \text{Number of moles} \]
\[ \text{Total Heat} = 13.7 \text{ kcal/mole} \times 1 \text{ mole} = 13.7 \text{ kcal} \]

### Conclusion

The value of heat generated when 36.5 g of HCl reacts with 40 g of NaOH during neutralization is **13.7 kcal**.

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