C2H5OH + 5O2 --> 3CO2 + 4H2O?

Given 212 g of C2H5OH and 115g of O2

a) calculate the mass of the water vapor that will be produced

b)Identify the limiting reactant

c)calculate the percentage yield if 68.0 g of water vapor is actually produced

e)calculate the mass of unused reactant.

Comments

  • The equation above^ seems unbalanced

    The # Oxygen on reactants is 11, and products is 10

    C2H5OH + 3O2 --> 2CO2 + 3H2O

    Given two masses from two reactants,

    there is bound to be a limiting reactant

    that runs out before the other... and

    the unused, or excess reactant.

    Grams to moles:

    212g x 1 mol C2H5OH/ 46.07g = 4.60 mol

    115g x 1 mol O2/32g = 3.59 mol

    Since molar ratio (coefficients)

    between C2H5OH and O2 is 1:3,

    4.6 x 3 = 13.8 moles O2 needed

    to completely combust

    But there's only 3.59 moles O2..

    so limiting reactant = O2

    a) Mass H2O (water vapour produced):

    Molar ratio O2 & H2O 3:3 = 1:1 same

    3.59 mol x 18.0g/mol = 64.6g produced

    b) O2

    c) 64.6g/ 68g x 100 = 95% yield

    e) 3.59/3 = 1.20 moles C2H5OH used

    4.60 - 1.20 = 3.40 moles excess

    3.40 mol x 46.07g/mol = 157g

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