Multiple choice

If LPG cylinder contains a mixture of butane and isobutane, then the amount of oxygen that would be required for combustion of $1$ kg of it will be:

  1. $1.8$ kg
  2. $2.7$ kg
  3. $4.5$ kg
  4. $3.58$ kg
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D Correct answer
Explanation

Combustion of butane (C4H10) and isobutane (C4H10) follows the same stoichiometry: 2C4H10 + 13O2 -> 8CO2 + 10H2O. 1 kg of C4H10 (MW 58) requires (13/2 * 32) / 58 kg of O2, which is approx 3.58 kg.

AI explanation

Both butane and isobutane share the same chemical formula of C4H10, so the complete combustion equation requires 13 moles of oxygen for every 2 moles of the gas. One mole of the gas has a mass of 58 g and requires 6.5 moles of oxygen, meaning 1 kg requires (1000 divided by 58) times 6.5 moles of oxygen. This equals 112.07 moles of oxygen, which multiplied by its molar mass of 32 g/mol gives a total required mass of approximately 3586 g or 3.58 kg.

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