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26 March, 01:52

The major source of aluminum in the world this bauxite (mostly aluminum oxide). It's thermal decomposition can be represented by:

Al2 O3 (s) - > 2 Al (s) + 3/2 O2 (g)

ΔH rxn = 1676

If aluminum is produced this way, how many grams of aluminum can conform when 1.000*10^3 kJ of heat is transferred?

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  1. 26 March, 06:01
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    The correct answer is 32.2 grams.

    Explanation:

    Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,

    ΔHrxn = 1676/2 = 838 kJ/mol

    Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,

    (1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum

    The grams of aluminum produced can be obtained by using the formula,

    mass = moles * molecular mass

    = 1.19 * 26.98

    = 32.2 grams.
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