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18 November, 10:58

Many grams of aluminum are required to produce 3.5 moles Al2O3 in the presence of excess O2?

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  1. 18 November, 12:23
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    The grams of aluminum that are required to produce 3.5 moles of AlO3 in presence of excess O2 is calculated as below

    write the equation for reaction

    4 Al + 3O2 = 2 Al2O3

    by use of mole ratio between Al to Al2O3 which is 4 : 2 the moles of Al

    =3.5 x4/2 = 7 moles

    mass of Al = moles / x molar mass

    = 7 moles x27 g/mol = 189 grams
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