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14 September, 01:05

An ore contains fe3o4 and no other iron. the iron in a 36.6-gram sample of the ore is all converted by a series of chemical reactions to fe2o3. the mass of fe2o3 is measured to be 29 g. what was the mass of fe3o4 in the sample of ore?

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  1. 14 September, 04:43
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    To solve this problem, let us first find for the molar mass of Fe2O3 and Fe3O4.

    Fe = 55.85 g/mol and O = 16 g/mol

    Therefore,

    Fe2O3 = 159.7 g/mol

    Fe3O4 = 231.55 g/mol

    We are given that there are 29 g of Fe2O3, we calculate for the amount of Fe from this in moles:

    mol Fe = 29 g Fe2O3 (1 / 159.7 g/mol) (2 mol Fe / 1 mol Fe2O3)

    mol Fe = 0.363 mol

    Converting this to Fe3O4:

    mass Fe3O4 = 0.363 mol Fe (1 mol Fe3O4 / 3 mol Fe) (231.55 g/mol)

    mass Fe3O4 = 28.03 g

    Therefore there are 28.03g of Fe3O4 in the ore.
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