The heat of combustion of ethane and benzene is `-1560` and `-3268" kJ mol"^(-1)` respectively. Which two has higher efficiency as fuel per gram and the amount of heat produced per gram?

Options

A. Benzene, `41.9" kJ g"^(-1)`
B. Ethane, `"52 kJ g"^(-1)`
C. Benzene, `"78 kJ g"^(-1)`
D. Ethane, `"30 kJ g"^(-1)`

Explanation

To determine which fuel, ethane or benzene, has higher efficiency per gram and produces more heat per gram, we will follow these steps:

### Step 1: Identify the heat of combustion values
- Heat of combustion of ethane = -1560 kJ/mol
- Heat of combustion of benzene = -3268 kJ/mol

### Step 2: Calculate the molar mass of each compound
- Molar mass of ethane (C2H6):
- Carbon (C): 12.01 g/mol × 2 = 24.02 g/mol
- Hydrogen (H): 1.008 g/mol × 6 = 6.048 g/mol
- Total molar mass of ethane = 24.02 + 6.048 = 30.07 g/mol (approximately 30 g/mol)

- Molar mass of benzene (C6H6):
- Carbon (C): 12.01 g/mol × 6 = 72.06 g/mol
- Hydrogen (H): 1.008 g/mol × 6 = 6.048 g/mol
- Total molar mass of benzene = 72.06 + 6.048 = 78.11 g/mol (approximately 78 g/mol)

### Step 3: Calculate the efficiency (heat of combustion per gram) for each fuel
- Efficiency of ethane:
\[ \text{Efficiency of ethane} = \frac{1560 \text{ kJ/mol}}{30 \text{ g/mol}} = 52 \text{ kJ/g} \]

- Efficiency of benzene:
\[ \text{Efficiency of benzene} = \frac{3268 \text{ kJ/mol}}{78 \text{ g/mol}} = 41.85 \text{ kJ/g} \approx 41.9 \text{ kJ/g} \]

### Step 4: Compare the efficiencies
- Ethane: 52 kJ/g
- Benzene: 41.9 kJ/g

### Conclusion
- Ethane has a higher efficiency as fuel per gram (52 kJ/g) compared to benzene (41.9 kJ/g).
- Therefore, ethane produces more heat per gram than benzene.

### Summary
- **Higher efficiency fuel per gram**: Ethane
- **Amount of heat produced per gram**: 52 kJ/g for ethane, 41.9 kJ/g for benzene.