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To solve this problem, we will calculate the enthalpy change per mole of the acid used in the reaction, using the given data.

### Step-by-Step Solution

#### Step 1: Calculate the Heat (q) Absorbed by the Solution
The heat absorbed (or released) by the solution can be calculated using the formula:
\[
q = m \cdot c \cdot \Delta T
\]
where:
- \( m \) is the total mass of the solution,
- \( c \) is the specific heat capacity of water (4.184 J/g°C),
- \( \Delta T \) is the change in temperature.

Given dataTo solve this problem, we will calculate the enthalpy change per mole of the acid used in the reaction, using the given data.

### Step-by-Step Solution

#### Step 1: Calculate the Heat (q) Absorbed by the Solution
The heat absorbed (or released) by the solution can be calculated using the formula:
\[
q = m \cdot c \cdot \Delta T
\]
where:
- \( m \) is the total mass of the solution,
- \( c \) is the specific heat capacity of water (4.184 J/g°C),
- \( \Delta T \) is the change in temperature.

Given 
- Volume of acid solution = 25.0 mL
- Volume of base solution = 25.0 mL
- Density of solutions = 1.00 g/mL, so the mass \( m \) is:
  \[
  m = 25.0 \, \text{mL} + 25.0 \, \text{mL} = 50.0 \, \text{g}
  \]

- Temperature change, \( \Delta T = 24.0^\circ \text{C} - 21.5^\circ \text{C} = 2.5^\circ \text{C} \)

Now, we can plug in the values:
\[
q = 50.0 \, \text{g} \times 4.184 \, \text{J/g°C} \times 2.5^\circ \text{C}
\]
\[
q = 522.75 \, \text{J}
\]

#### Step 2: Calculate the Moles of the Limiting Reactant (HCl)
The molarity of the strong acid (HCl) solution is given as 1.2 M, and the volume is 25.0 mL (or 0.025 L).

Using the formula:
\[
\text{moles of HCl} = \text{Molarity} \times \text{Volume}
\]
\[
\text{moles of HCl} = 1.2 \, \text{mol/L} \times 0.025 \, \text{L}
\]
\[
\text{moles of HCl} = 0.03 \, \text{mol}
\]

#### Step 3: Calculate the Enthalpy Change per Mole of Acid
The enthalpy change per mole of acid (\( \Delta H \)) can be found by dividing the heat \( q \) by the moles of HCl. Since the reaction is exothermic, the enthalpy change will be negative.

\[
\Delta H = \frac{q}{\text{moles of HCl}}
\]
\[
\Delta H = \frac{522.75 \, \text{J}}{0.03 \, \text{mol}}
\]
\[
\Delta H = 17425 \, \text{J/mol}
\]

To convert this into kJ/mol:
\[
\Delta H = 17.4 \, \text{kJ/mol}
\]

### Answer
The enthalpy change per mole of acid is approximately **-17.4 kJ/mol** (negative because heat is released in the reaction).

---

Would you like further details or have questions about this calculation?

### Related Questions:
1. How would the result change if the density of the solution was different?
2. What would happen to \( \Delta H \) if we doubled the volume of both acid and base?
3. How would the enthalpy change be affected if a weak acid was used instead?
4. Why is the heat absorbed considered as the enthalpy change of the reaction?
5. How could we measure the enthalpy change in a laboratory setting with more accuracy?

### Tip:
Always double-check units when calculating enthalpy changes to avoid errors, especially when converting from Joules to kilojoules.

If 25.0 mL of a strong acid with a concentration of 1.2 M is mixed with 25.0 mL of a strong base, and the temperature rises from 21.5°C to 24.0°C, calculate the enthalpy change per mole of acid.

Find the enthalpy change per mole of a strong acid in a reaction with a strong base, using given temperature rise and concentration. This solution involves concepts in thermochemistry, stoichiometry, and enthalpy, and includes step-by-step calculations using q = m * c * ΔT and ΔH = q / moles formulas.

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