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8 March, 06:33

Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 4.3 g of hexane is mixed with 7.14 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.

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  1. 8 March, 06:54
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    We can produce 6.20 grams of CO2

    Explanation:

    Step 1: Data given

    Mass of hexane = 4.3 grams

    Molar mass of hexane = 86.18 g/mol

    Mass of oxygen = 7.14 grams

    Molar mass of oxygen = 32.0 g/mol

    Step 2: The balanced equation

    2C6H14 + 19O2 → 12CO2 + 14H2O

    Step 3: Calculate moles

    Moles = mass / molar mass

    Moles hexane = 4.3 grams / 86.18 g/mol

    Moles hexane = 0.0499 moles

    Moles oxygen = 7.14 grams / 32.0 g/mol

    Moles oxygen = 0.2231 moles

    Step 4: Calculate the limiting reactant

    For 2 moles hexane we need 19 moles O2 to produce 12 moles CO2 and 14 moles H2O

    Oxygen is the limiting reactant. It will completely be consumed (0.2231 moles). Hexane is in excess. There will react 2/19 * 0.2231 = 0.02348 moles

    There will be porduced 12/19 * 0.2231 = 0.1409 moles CO2

    Step 5: Calculate mass CO2

    Mass CO2 = moles CO2 * molar mass CO2

    Mass CO2 = 0.1409 moles * 44.01 g/mol

    Mass CO2 = 6.20 grams

    We can produce 6.20 grams of CO2
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