The heat of combustion of ethane gas at 373 k cal per mole. Assume that 50% of heat is lost, how many litres of ethane measured at STP must be convert `50` kg of water at `10^(@)C` to steam at `100^(@)C` ? One mole of gas oc cupies `22.4` litres at STP. Take latent heat of steam `=2.25xx10^(6) j kg^(-1)` .

Explanation

To solve the problem, we need to find out how many liters of ethane gas are required to convert 50 kg of water at 10°C to steam at 100°C, considering that 50% of the heat produced by the combustion of ethane is lost.

### Step 1: Calculate the total heat required to convert water at 10°C to steam at 100°C.

1. **Heat required to raise the temperature of water from 10°C to 100°C:**
- The specific heat capacity of water is approximately \(4.18 \, \text{J/g°C}\).
- Mass of water = 50 kg = 50000 g
- Temperature change = \(100°C - 10°C = 90°C\)

\[ Q_1 = m \cdot c \cdot \Delta T = 50000 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot 90°C \]

\[ Q_1 = 50000 \cdot 4.18 \cdot 90 = 188100000 \, \text{J} \]

2. **Heat required to convert water at 100°C to steam:**
- Latent heat of steam = \(2.25 \times 10^6 \, \text{J/kg}\)
- Mass of water = 50 kg

\[ Q_2 = m \cdot L = 50 \, \text{kg} \cdot 2.25 \times 10^6 \, \text{J/kg} \]

\[ Q_2 = 112500000 \, \text{J} \]

3. **Total heat required (Q_total):**

\[ Q_{\text{total}} = Q_1 + Q_2 = 188100000 \, \text{J} + 112500000 \, \text{J} = 300600000 \, \text{J} \]

### Step 2: Calculate the effective heat produced by the combustion of ethane.

1. **Heat of combustion of ethane = 373 kcal/mole.**
- Convert kcal to Joules (1 kcal = 4184 J):

\[ \text{Heat of combustion in J} = 373 \, \text{kcal} \times 4184 \, \text{J/kcal} = 1550000 \, \text{J/mole} \]

2. **Considering 50% heat loss:**

\[ \text{Effective heat} = 0.5 \times 1550000 \, \text{J/mole} = 775000 \, \text{J/mole} \]

### Step 3: Calculate the number of moles of ethane required.

1. **Number of moles required (n):**

\[ n = \frac{Q_{\text{total}}}{\text{Effective heat}} = \frac{300600000 \, \text{J}}{775000 \, \text{J/mole}} \approx 387.74 \, \text{moles} \]

### Step 4: Calculate the volume of ethane at STP.

1. **Volume of ethane (V):**
- One mole of gas occupies 22.4 liters at STP.

\[ V = n \cdot 22.4 \, \text{liters/mole} = 387.74 \, \text{moles} \cdot 22.4 \, \text{liters/mole} \approx 8694.66 \, \text{liters} \]

### Final Answer:
Approximately **8694.66 liters** of ethane are required to convert 50 kg of water at 10°C to steam at 100°C.

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