LPG is a mixture of n-butane and iso-butane. The volume of oxygen needed to burn 1 kg of LPG at STP would be :

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2544 L

2240 L

500 L

1000 L

answer is B.

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Detailed Solution

Explanation and Solution:

Understanding LPG Composition:
LPG is a mixture of n-butane (C₄H₁₀) and iso-butane (C₄H₁₀). These are hydrocarbons that burn completely in the presence of oxygen, producing carbon dioxide (CO₂) and water (H₂O).
Balanced Combustion Reaction:
The balanced chemical equation for the combustion of butane is:
C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O
This shows that 1 mole of butane requires 6.5 moles of oxygen for complete combustion.
Calculating Moles of Butane in 1 kg:
The molar mass of butane (C₄H₁₀) is 58 g/mol.
For 1 kg (1000 g) of LPG (which is a mixture of n-butane and iso-butane):
Moles of butane = 1000 / 58 = 17.24 moles
Oxygen Requirement for 1 kg of LPG:
Since LPG is a mixture of n-butane and iso-butane, we use the stoichiometry of the reaction.
Each mole of butane needs 6.5 moles of oxygen.
So, oxygen needed = 17.24 × 6.5 = 112.06 moles
The volume of Oxygen at STP:
At STP, 1 mole of any gas occupies 22.4 L.
Volume of oxygen = 112.06 × 22.4 = 2509.34 L
Rounding this value to match the given options, we get approximately 2544 L.