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26 March, 21:21

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 9.32 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. Round your answer to significant digits

2 C2H6 + 7 O2 = 4 CO2 + 6 H2O

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  1. 27 March, 00:16
    0
    There will remain 8.06 grams of ethane

    Explanation:

    Step 1: Data given

    Mass of ethane = 9.32 grams

    Mass of oxygen = 12.0 grams

    Molar mass ethane = 30.07 g/mol

    Molar mass oxygen = 32.00 g/mol

    Step 2: The balanced equation

    2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

    Step 3: Calculate moles ethane

    Moles ethane = mass ethane / molar mass ethane

    Moles ethane = 9.32 grams / 30.07 g/mol

    Moles ethane = 0.3099 moles

    Step 4: Calculate moles oxygen

    Moles oxygen = 12.0 grams / 32.0 g/mol

    Moles oxygen = 0.375 moles

    Step 5: Calculate the limiting reactant

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

    O2 is the limiting reactant. It will completely be consumed (0.375 moles)

    Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

    There will remain 0.375 - 0.107 = 0.268 moles

    Step 6: Calculate mass ethane

    Mass ethane = moles ethane * molar mass ethane

    Mass ethane = 0.268 moles * 30.07 g/mol

    Mass ethane = 8.06 grams

    There will remain 8.06 grams of ethane
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