How much heat is liberated when one mole of gaseous `Na^(o+)` combines with one mole of `Cl^(Theta)` ion to form solid `NaCl`. Use the data given below:
`{:(Na(s)+(1)/(2)Cl_(2)(g)rarrNaCl(s),,DeltaH =- 98.23 kcal),(Na(s)rarrNa(g),,DeltaH =+25.98 kcal),(Na(g)rarrNa^(o+)+e^(-),,DeltaH =+120.0 kcal),(Cl_(2)(g)rarr2Cl(g),,DeltaH =+58.02 kcal),(Cl^(Theta)(g)rarrCl(g)+e^(-),,DeltaH=+87.3kcal):}`

Explanation

To determine how much heat is liberated when one mole of gaseous \( \text{Na}^+ \) combines with one mole of \( \text{Cl}^- \) to form solid \( \text{NaCl} \), we will use Hess's law. We will rearrange the given reactions and their enthalpy changes to derive the desired reaction.

### Step-by-Step Solution:

1. **Write down the given reactions and their enthalpy changes:**
- \( \text{Na(s)} + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{NaCl(s)} \), \( \Delta H = -98.23 \, \text{kcal} \)
- \( \text{Na(s)} \rightarrow \text{Na(g)} \), \( \Delta H = +25.98 \, \text{kcal} \)
- \( \text{Na(g)} \rightarrow \text{Na}^+(g) + e^- \), \( \Delta H = +120.0 \, \text{kcal} \)
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