Question

What mass of iron will be obtained in pure state from 100kg of an iron ore containing 90% Fe2O3 ? ( Fe -56 amu )

Answer 1

wt of Fe2O3 = 90 kg
wt of fe = 2*56*90/160=63 kg

Answer 2

In production of iron (Fe) for use in iron and steel industries, Iron (III) oxide (Fe₂O₃) is a very important mineral ore (hematite).

Iron (III) oxide is used in many industrial reactions such as carbothermal reduction to produce iron.

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The mass of FeO3 in the iron ore will be:

90% of the 100kg ore

90/100 × 100   =  90 kg of Fe₂O₃

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The carbothermal reaction for the reduction of Fe₂O₃ to Fe is as follows:

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

The mole ratio between Fe₂O₃: Fe  is 1:2

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Find moles of 90 kg Fe₂O₃

moles = mass/molar mass

mass= 90000 g
molar mass = 160

moles = 90000/160
          =  562.5 moles

The mole ratio between Fe₂O₃ and Fe in the equation above was 1:2

If 1 mole of Fe₂O₃ will give 2 moles of Fe
Then 562.5 moles  = 562.5 × 2/1

                               = 1125 moles

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Find mass of Fe that corresponds to 1125 moles

mass = moles × molar mass

        =   1125 × 56

       = 63000 g

      = 63 Kg

The amount of iron liberated will be 63kg
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