The heat evolved from the combustion of carbon is used to heat water. Assuming `50%` efficiency, calculate mole of water vaporized at its boiling point `Delta H_(f)=(CO_(2))=-94K cal//mol` and `DeltaH_(vap)(H_(2)O)=9.6kcal //mol)` and `6g C` is undergoing combustion

Options

A. `1.21`mole
B. 2.42 mole
C. 4.89 mole
D. 9.7 mole

Explanation

To solve the problem step by step, we will follow the given information and perform the necessary calculations.

### Step 1: Calculate the moles of carbon combusted
Given:
- Mass of carbon (C) = 6 g
- Molar mass of carbon = 12 g/mol

To find the number of moles of carbon:
\[ \text{Number of moles of C} = \frac{\text{mass of C}}{\text{molar mass of C}} = \frac{6 \, \text{g}}{12 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 2: Calculate the heat released from the combustion of carbon
Given:
- \(\Delta H_f (CO_2) = -94 \, \text{kcal/mol}\)

Since 0.5 moles of carbon are combusted, the total heat released (without considering efficiency) can be calculated as:
\[ \text{Heat released} = \text{moles of C} \times \Delta H_f = 0.5 \, \text{mol} \times (-94 \, \text{kcal/mol}) = -47 \, \text{kcal} \]

### Step 3: Adjust for efficiency
Given that the efficiency is 50%, the effective heat available for vaporizing water is:
\[ \text{Effective heat} = \frac{-47 \, \text{kcal}}{2} = -23.5 \, \text{kcal} \]

### Step 4: Calculate the moles of water vaporized
Given:
- \(\Delta H_{vap}(H_2O) = 9.6 \, \text{kcal/mol}\)

Let \( n \) be the number of moles of water vaporized. The heat required to vaporize \( n \) moles of water is:
\[ \text{Heat required} = n \times \Delta H_{vap} = n \times 9.6 \, \text{kcal/mol} \]

Setting the heat released equal to the heat required for vaporization:
\[ -23.5 \, \text{kcal} = n \times 9.6 \, \text{kcal/mol} \]

### Step 5: Solve for \( n \)
\[ n = \frac{-23.5 \, \text{kcal}}{9.6 \, \text{kcal/mol}} \approx 2.44 \, \text{mol} \]

### Conclusion
The number of moles of water vaporized at its boiling point is approximately \( 2.44 \, \text{mol} \).

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