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13 December, 18:56

If you burn 55.6 g of hydrogen and produce 497 g of water, how much oxygen reacted?

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Answers (2)
  1. 13 December, 21:13
    0
    2H2 + O2 → 2H2O

    2 mol H2 + 1mol O2 will produce 2 mol H2O

    Molar mass H2O = 18.0153g/mol

    497g H2O = 23.813 mol H2O produces

    This will require 27.587/2 = 13.793 mol O2

    Molar mass O2 = 32g/mol

    13.793mol = 13.793*32 = 441.37g O2 required.

    The easiest way is subtracting 55.6 from 497 which gives you 441.4.
  2. 13 December, 22:11
    0
    441.28 g Oxygen

    Explanation:

    The combustion of hydrogen gives water as the product. The equation for the reaction is;

    2H₂ (g) + O₂ (g) → 2H₂O (l)

    Mass of hydrogen = 55.6 g

    Number of moles of hydrogen

    Moles = Mass/Molar mass

    = 55.6 g : 2.016 g/mol

    = 27.8 moles

    The mole ratio of Hydrogen to Oxygen is 2:1

    Therefore;

    Number of moles of oxygen = 27.5794 moles : 2

    = 13.790 moles

    Mass of oxygen gas will therefore be;

    Mass = Number of moles * Molar mass

    Molar mass of oxygen gas is 32 g/mol

    Mass = 13.790 moles * 32 g/mol

    = 441.28 gAlternatively:

    Mass of hydrogen + mass of oxygen = Mass of water

    Therefore;

    Mass of oxygen = Mass of water - mass of hydrogen

    = 497 g - 55.6 g

    = 441.4 g
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