A cylinder of cooking gas is assumed to contain 11.2 kg of butane. The thermo chemical equation for the combustion of butane is : `C_(4)H_(10)(g)+(13)/(2)O_(2)(g)rarr 4CO_(2)(g)+5H_(2)O(l) , Delta H = -2658 kJ`
If a family needs 15000 kJ of energy per day for cooking, how would the cylinder last ?

Explanation

To solve the problem, we need to determine how long a cylinder containing 11.2 kg of butane will last if a family requires 15,000 kJ of energy per day for cooking. We will follow these steps:

### Step 1: Calculate the number of moles of butane (C₄H₁₀)
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)}} \]
First, convert the mass of butane from kg to grams:
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