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13 April, 01:22

Gaseous ethane (CH, CH) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 24 g of

ethane is misced with 16.9 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2

significant digits.

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Answers (1)
  1. 13 April, 01:26
    0
    Mass = 8.1 g

    Explanation:

    Given dа ta:

    Mass of ethane = 24 g

    Mass of oxygen = 16.9 g

    Mass of water = ?

    Solution:

    Chemical equation:

    2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

    Moles of ethane:

    Number of moles = mass/molar mass

    Number of moles = 24 g / 30.07 g/mol

    Number of moles = 0.8 mol

    Moles of oxygen:

    Number of moles = mass/molar mass

    Number of moles = 16.9 g / 32 g/mol

    Number of moles = 0.53 mol

    Now we will compare the moles of water with oxygen and ethane.

    C₂H₆ : H₂O

    2 : 6

    0.8 : 6/2*0.8 = 2.4 mol

    O₂ : H₂O

    7 : 6

    0.53 : 6/7*0.53 = 0.45 mol

    It means oxygen is limiting reactant because moles of water produced by given amount of oxygen are less.

    Mass of water:

    Mass = number of moles * molar mass

    Mass = 0.45 mol * 18 g/mol

    Mass = 8.1 g
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