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8 May, 19:18

If 25.4 grams of water react to form 2.8 grams of hydrogen, how many grams of oxygen must simultaneously be formed?

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Answers (2)
  1. 8 May, 19:43
    0
    There will be formed 22.6 grams O2

    Explanation:

    Step 1: Data given

    Mass of 25.4 grams

    Mass of hydrogen formed = 2.8 grams

    Step 2: The balanced equation

    2H2O (l) → 2H2 (aq) + O2 (aq)

    For 2 moles H2O we'll have 2 moles hydrogen and 1 mol O2

    Step 3: Calculate moles of H2O

    Moles H2O = mass H2O / molar mass H2O

    Moles H2O = 25.4 grmas / 18.02 g/mol

    Moles H2O = 1.41 moles

    Step 4: Calculate moles hydrogen formed

    Moles H2 = 2.8 grams / 2.0 g/mol

    Moles H2 = 1.4 moles

    Step 5: Calculate moles of O2

    For 2 mole of H2O reacted, there will be formed 1 mol of H2O

    For 1.41 moles of H2O formed there will be formed 0.705moles O2.

    Step 6: Calculate mass O2

    Mass O2 = moles O2 * molar mass O2

    Mass O2 = 0.705 * 32 g/mol

    Mass O2 = 22.6 grams O2

    There will be formed 22.6 grams O2
  2. 8 May, 20:47
    0
    The mass of oxygen and hydrogen must be equal to the mass of the substance they create the water. So if the hydrogen is 2.8 g the oxygen must account for the rest of the mass. Basically just subtract 25.4-2.8=mass of oxygen
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