a. A cylinder of gas is assumed to contain `11.2 kg` of butane. If a normal family needs `20000kJ` of energy per day for cooking, how long will the cylinder last? Given that the heat of combustion of butane is `2658 kJ mol^(-1)`.
b. If the air supply of the burner is insufficient (i.e. you have a yellow instead of a blue flame), a portion of the gas escape without combustion. Assuming that `33%` of the gas is wasted due to this inefficiency, how long would the cylinder last?

Options

A. 20 days
B. 25 days
C. 28 days
D. 24 days

Answer

To solve the given problem, we will break it down into two parts as specified in the question.

### Part A:
1. **Convert the mass of butane from kilograms to grams**:
\[ \text{Mass of butane} = 11.2 \, \text{kg} = 11.2 \times 1000 \, \text{g} = 11200 \, \text{g} \]

2. **Calculate the molar mass of butane (C₄H₁₀)**:
- Carbon (C) = 12 g/mol, and there are 4 carbon atoms.
- Hydrogen (H) = 1 g/mol, and there are 10 hydrogen atoms.
\[ \text{Molar mass of butane} = (4 \times 12) + (10 \times 1) = 48 + 10 = 58 \, \text{g/mol} \]

3. **Calculate the number of moles of butane**:
\[ \text{Number of moles of butane} = \frac{\text{Mass of butane}}{\text{Molar mass of butane}} = \frac{11200 \, \text{g}}{58 \, \text{g/mol}} \approx 193.10 \, \text{mol} \]

4. **Calculate the total energy released from the combustion of butane**:
\[ \text{Energy released} = \text{Number of moles} \times \text{Heat of combustion} \]
\[ \text{Energy released} = 193.10 \, \text{mol} \times 2658 \, \text{kJ/mol} \approx 513268.8 \, \text{kJ} \]

5. **Calculate how many days the cylinder will last**:
\[ \text{Daily energy requirement} = 20000 \, \text{kJ/day} \]
\[ \text{Number of days} = \frac{\text{Total energy released}}{\text{Daily energy requirement}} = \frac{513268.8 \, \text{kJ}}{20000 \, \text{kJ/day}} \approx 25.66 \, \text{days} \]
Rounding this gives approximately **26 days**.

### Part B:
1. **Calculate the effective energy utilized considering the inefficiency**:
- If 33% of the gas is wasted, then 67% is utilized.
\[ \text{Effective energy utilized} = 513268.8 \, \text{kJ} \times \left(1 - 0.33\right) = 513268.8 \, \text{kJ} \times 0.67 \approx 343890.0 \, \text{kJ} \]

2. **Calculate how many days the cylinder will last with the inefficiency**:
\[ \text{Number of days with inefficiency} = \frac{343890.0 \, \text{kJ}}{20000 \, \text{kJ/day}} \approx 17.19 \, \text{days} \]
Rounding this gives approximately **17 days**.

### Final Answers:
- Part A: The cylinder will last approximately **26 days**.
- Part B: The cylinder will last approximately **17 days**.