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3 September, 11:53

Mining companies extract iron from iron ore according to the following balanced equation: Fe2O3 (s) + 3CO (g) →2Fe (s) + 3CO2 (g) In a reaction mixture containing 175 g Fe2O3 and 73.0 g CO, CO is the limiting reactant. Calculate the mass of the reactant in excess (which is Fe2O3) that remains after the reaction has gone to completion. Express the mass with the appropriate units.

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  1. 3 September, 14:45
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    There will remain 36.7 grams of Fe2O3

    Explanation:

    Step 1: Data given

    Mass of Fe2O3 = 175 grams

    Mass of CO = 73.0 grams

    CO is the limiting reactant

    Molar mass of CO = 28 g/mol

    Molar mass of Fe2O3 = 159.69 g/mol

    Step 2: The balanced equation

    Fe2O3 (s) + 3CO (g) →2Fe (s) + 3CO2 (g)

    Step 3: Calculate moles of CO

    Moles CO = mass CO / molar mass CO

    Moles CO = 73.0 grams / 28.0 g/mol

    Moles CO = 2.61 moles

    Step 4: Calculate moles of Fe2O3

    Moles Fe2O3 = 175.0 grams / 159.69 g/mol

    moles Fe2O3 = 1.10 moles

    Step 5: Calculate moles reacted

    Co is the limiting reactant. It will completely be consumed (2.61 moles)

    Fe2O3 is in excess. There will be 2.61/3 = 0.87 moles of Fe2O3 consumed

    There will remain 1.10 - 0.87 = 0.23 moles of Fe2O3

    Step 6: Calculate the mass of Fe2O3 that remains

    Mass Fe2O3 = moles * molar mass

    Mass Fe2O3 = 0.23 moles * 159.69 g/mol

    Mass of Fe2O3 = 36.7 grams

    There will remain 36.7 grams of Fe2O3
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