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Consider the reaction between HCl and O2: 4HCl(g)+O2(g)→2H2O(l)+2Cl2(g)?

When 63.1 g of HCl is allowed to react with 17.2 g of O2, 54.4 g of Cl2 is collected.

Determine the theoretical yield of Cl2 for this reaction.        

2 Answers

Relevance
  • 1 month ago

    63.1 g HCl / 36.46 g/mol X (2 mol Cl2 / 4 mol HCl) X 70.9 g/mol = 61.4 g Cl2

    17.2 g O2 / 32.0 g/mol X (2 mol Cl2/1 mol O2) X 70.9 g/mol = 135 g Cl2

    HCl is the limiting reactant, and the theoretical yield of Cl2 is 61.4 g

  • 1 month ago

    4 HCl + O2 → 2 H2O + 2 Cl2

    (63.1 g HCl) / (36.4611 g HCl/mol) = 1.73061 mol HCl

    (17.2 g O2) / (31.99886 g O2/mol) = 0.537519 mol O2

    1.73061 moles of HCl would react completely with 1.73061 x (1/4) =

    0.4326525 mole of O2, but there is more O2 present than that, so O2 is in excess and HCl is the limiting reactant.

    (1.73061 mol HCl) x (2 mol Cl2 / 4 mol HCl) x (70.9064 g Cl2/mol) = 61.356 g =

    61.4 g Cl2 in theory

    - - - - - - - - - - - - - - - -

    And anticipating your next question:

    (54.4 g) / (61.4 g) = 0.886 = 88.6% yield Cl2

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