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13 May, 13:10

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 2.10 g of ethane is mixed with 12. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

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  1. 13 May, 13:50
    0
    Since ethane is the limting reactant, there will remain nothing. The mass ethane remaingin is 0 grams

    There remains 4.18 grams of oxygen

    Explanation:

    Step 1: data given

    Ethane = C2H6

    Mass of ethane = 2.10 grams

    Molar mass C2H6 = 30.07 g/mol

    oxygen = O2

    Mass of oxygen = 12.0 grams

    Molar mass O2 = 32.0 g/mol

    Step 2: The balanced equation

    2C2H6 + 7O2 → 4CO2 + 6H2O

    Step 3: Calculate moles C2H6

    Moles C2H6 = mass C2H6 / molar mass C2H6

    Moles C2H6 = 2.10 grams / 30.07 g/mol

    Moles C2H6 = 0.0698 moles

    Step 4: Calculate moles O2

    Moles O2 = 12.0 grams / 32.0 g/mol

    Moles O2 = 0.375 moles

    Step 5: Calculate limiting reactant

    For 2 moles ethane we have 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

    Ethane is the limiting reactant. It wil completely be consumed (0.0698 moles). O2 is in excess. There will react 3.5 * 0.0698 = 0.2443 moles

    There will remain 0.375 - 0.2443 = 0.1307 moles

    Since ethane is the limting reactant, there will remain nothing. The mass ethane remaingin is 0 grams

    Step 6: Calculate mass oxygen remaining

    Mass oxygen = 0.1307 moles * 32.0 g/mol

    Mass oxygen remaining = 4.18 grams
  2. 13 May, 14:10
    0
    Ethane is the limiting reactant, so no mass of ethane will be left over by the chemical reaction. All the mass will react, the 2.1 grams of ethane.

    Explanation:

    First of all, we need to determine the reaction and the limiting reactant to work with the stoichiometry.

    The equation is: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

    We define the moles of the reactants:

    2.10 g / 30 g/mol = 0.07 moles of ethane

    12 g / 32 g/mol = 0.375 moles of oxygen

    To determine the limiting reactant, we start with oxygen:

    7 moles of O₂ can react with 2 moles of ethane

    Then, 0.375 moles of O₂ will react with (0.375. 2) / 7 = 1.31 moles of ethane.

    We do not have enough ethane, just only 0.07 moles to react.

    Ethane is the limiting reactant, so no mass of ethane will be left over by the chemical reaction. All the mass will react, the 2.1 grams of ethane.
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