The composition of LPG is butane and isobutane. The amount of oxygen that would be required for combustion of 1 kg of LPG will be approximately.

Options

A. 4.5 Kg
B. 1.8 Kg
C. 2.7 Kg
D. 3.6 Kg

Answer

To determine the amount of oxygen required for the combustion of 1 kg of LPG (which is a mixture of butane and isobutane), we can follow these steps:

### Step 1: Understand the Composition of LPG
LPG primarily consists of two isomers: butane (C4H10) and isobutane (C4H10). Both have the same molecular formula but different structures.

### Step 2: Write the Combustion Reaction
The balanced chemical equation for the complete combustion of butane is:
\[ 2 \, \text{C}_4\text{H}_{10} + 13 \, \text{O}_2 \rightarrow 8 \, \text{CO}_2 + 10 \, \text{H}_2\text{O} \]

### Step 3: Calculate the Molar Masses
- Molar mass of butane (C4H10):
\[ \text{Molar mass} = (4 \times 12) + (10 \times 1) = 48 + 10 = 58 \, \text{g/mol} \]
- Molar mass of oxygen (O2):
\[ \text{Molar mass} = 2 \times 16 = 32 \, \text{g/mol} \]

### Step 4: Determine the Amount of Oxygen Required
From the balanced equation, we see that:
- 2 moles of butane (116 g) require 13 moles of oxygen (416 g).

### Step 5: Calculate the Oxygen Required for 1 kg of LPG
We need to find out how much oxygen is required for 1000 g (1 kg) of LPG:
\[ \text{Oxygen required} = \left( \frac{416 \, \text{g O}_2}{116 \, \text{g C}_4\text{H}_{10}} \right) \times 1000 \, \text{g} \]
Calculating this gives:
\[ \text{Oxygen required} = \frac{416}{116} \times 1000 \approx 3586.2 \, \text{g} \]

### Step 6: Convert to Kilograms
To convert grams to kilograms:
\[ \text{Oxygen required in kg} = \frac{3586.2}{1000} \approx 3.586 \, \text{kg} \]

### Step 7: Round Off the Result
Rounding off 3.586 kg gives approximately 3.6 kg.

### Final Answer
The amount of oxygen required for the combustion of 1 kg of LPG is approximately **3.6 kg**.

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