The amount of heat released during the reaction of `300mL` of `1M HCl` with ``100mL` of `1M NaOH` is `:`

Options

A. `573kJ`
B. `573xx10^(-3)Kj`
C. `57.3kJ`
D. `5.73kJ`

Explanation

To find the amount of heat released during the reaction of 300 mL of 1M HCl with 100 mL of 1M NaOH, we can follow these steps:

### Step 1: Calculate the number of moles of HCl and NaOH
- **HCl**:
- Volume = 300 mL = 0.300 L
- Molarity = 1 M
- Moles of HCl = Molarity × Volume = 1 mol/L × 0.300 L = 0.300 moles

- **NaOH**:
- Volume = 100 mL = 0.100 L
- Molarity = 1 M
- Moles of NaOH = Molarity × Volume = 1 mol/L × 0.100 L = 0.100 moles

### Step 2: Determine the limiting reactant
- The reaction between HCl and NaOH is a 1:1 reaction:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- We have 0.300 moles of HCl and 0.100 moles of NaOH. Since NaOH is the limiting reactant, it will determine the amount of heat released.

### Step 3: Calculate the heat released during the neutralization
- The heat of neutralization for strong acids and bases is typically around -57.27 kJ for 1000 mEq (or 1 mole).
- Since we have 0.100 moles of NaOH (which is equivalent to 100 mEq), we can calculate the heat released:
\[ \text{Heat released} = \left( \frac{57.27 \text{ kJ}}{1000 \text{ mEq}} \right) \times 100 \text{ mEq} = 5.727 \text{ kJ} \]

### Final Answer:
The amount of heat released during the reaction is **5.727 kJ**.

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