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)`.

Explanation

To solve the problem step by step, we need to find out how long a cylinder containing 11.2 kg of butane will last if a family requires 20,000 kJ of energy per day for cooking, given that the heat of combustion of butane is 2658 kJ/mol.

### Step 1: Determine the molar mass of butane
The molecular formula for butane is C₄H₁₀. The molar mass can be calculated as follows:
- Carbon (C): 12.01 g/mol × 4 = 48.04 g/mol
- Hydrogen (H): 1.008 g/mol × 10 = 10.08 g/mol
- Total molar mass of butane = 48.04 g/mol + 10.08 g/mol = 58.12 g/mol (approximately 58 g/mol)

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