In the oxidation of iron; how many grams of iron (III) oxide will be produced from 6.20 mol of Fe? 4Fe(s) + 3O2(g) –> 2Fe2O3(s)

Question

In the oxidation of iron; how many grams of iron (III) oxide will be produced from 6.20 mol of Fe?
4Fe(s) + 3O2(g) –> 2Fe2O3(s)

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Thiên Ân 5 months 2021-09-01T03:33:47+00:00 1 Answers 2 views 0

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    2021-09-01T03:35:40+00:00

    Answer:

    496 g of Fe₂O₃.

    Explanation:

    The balanced equation for the reaction is given below:

    4Fe + 3O₂ —> 2Fe₂O₃

    From the balanced equation above,

    4 moles of Fe reacted to produce 2 moles of Fe₂O₃.

    Therefore, 6.20 moles of Fe will react to produce = (6.20 × 2)/4 = 3.1 moles of Fe₂O₃

    Finally, we shall determine the mass of 3.1 moles of Fe₂O₃. This can be obtained as follow:

    Mole of Fe₂O₃ = 3.1 moles

    Molar mass of Fe₂O₃ = (56 × 2) + (3×16)

    = 112 + 48

    = 160 g/mol

    Mass of Fe₂O₃ =?

    Mass = mole × molar mass

    Mass of Fe₂O₃ = 3.1 × 160

    Mass of Fe₂O₃ = 496 g

    Therefore, 496 g of Fe₂O₃ were produced from the reaction.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )