A cylinder of gas is assumed to contain 11.6 kg of butane `(C_4H_10)`. If a normal family needs 20000kJ of energy per day, then the cylinder will last for : (Given that ΔH for combustion of butane is – 2658kJ)

Options

A. 25.66 days
B. 31 days
C. 28.5 days
D. 22.0 days

Explanation

To solve the problem, we need to determine how long a cylinder containing 11.6 kg of butane will last, given that a family requires 20,000 kJ of energy per day and the heat of combustion of butane is -2658 kJ per mole.

### Step-by-Step Solution:

1. **Calculate the Molar Mass of Butane (C₄H₁₀)**:
- The molar mass of carbon (C) = 12 g/mol
- The molar mass of hydrogen (H) = 1 g/mol
- Molar mass of butane (C₄H₁₀) = (4 × 12) + (10 × 1) = 48 + 10 = 58 g/mol

2. **Convert the Mass of Butane from kg to g**:
- Given mass of butane = 11.6 kg
- Convert to grams: 11.6 kg × 1000 g/kg = 11600 g

3. **Calculate the Number of Moles of Butane**:
- Number of moles = mass (g) / molar mass (g/mol)
- Number of moles of butane = 11600 g / 58 g/mol = 200 moles

4. **Calculate the Total Energy Released by the Combustion of Butane**:
- Energy released per mole of butane = -2658 kJ
- Total energy released = number of moles × energy per mole
- Total energy = 200 moles × 2658 kJ/mole = 531600 kJ

5. **Determine How Many Days the Cylinder Will Last**:
- Daily energy requirement = 20,000 kJ/day
- Number of days the cylinder will last = total energy / daily energy requirement
- Days = 531600 kJ / 20000 kJ/day = 26.58 days

### Final Answer:
The cylinder will last for approximately **26.58 days**.

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