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The amount (in liters) of Oxygen at STP that is required for the combustion of 4gm of ethylene is :

Correct answer: B. 9.6 liters

  • A. 96 liters
  • B. 9.6 liters
  • C. 44.8 liters
  • D. 7.2 liters

Explanation

Your reaction will be C2H4 + 3O2 -> 2CO2 + 2H2O 4g of ethylene is 1/7 mole 1 ethylene reacts with 3O2 1/7 will react with x X= 3/7 moles At stp so moles=volume/22.4 Volume= moles× 22.4 9.6 litres

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About Fundamental Concepts of Chemistry

Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.

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