Calculate heat of formation of `KOH(s)` using the following equations
`K(s) +H_(2)O(l) +aq rarr KOH(aq)+1//2 H_(2)O(g),DeltaH =- 48.0 kcal …(i)`
`H_(2)(g) +2O_(2)(g) rarr H_(2)O(l),DeltaH =- 68.4 kcal …(ii)`
`KOH(s) +(aq) rarr KOH(aq),DeltaH =- 14.0 kcal ....(iii)`

Explanation

To calculate the heat of formation of KOH(s) using the given equations, we will follow a systematic approach. The heat of formation of a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states.

### Given Equations:
1. \( K(s) + H_2O(l) \rightarrow KOH(aq) + \frac{1}{2} H_2O(g), \Delta H = -48.0 \, \text{kcal} \) (i)
2. \( H_2(g) + 2O_2(g) \rightarrow H_2O(l), \Delta H = -68.4 \, \text{kcal} \) (ii)
3. \( KOH(s) + (aq) \rightarrow KOH(aq), \Delta H = -14.0 \, \text{kcal} \) (iii)

### Step-by-Step Solution:
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