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 :

Explanation and Solution:

1. 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).
2. 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.
3. 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
4. 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
5. 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
6. Rounding this value to match the given options, we get approximately 2544 L.

Final Answer:

b) 2544 L