# Iron (III) oxide may be reduced to pure iron either with carbon monoxide or with coke (pure carbon). Suppose that 150kg of ferric oxide are available. How many kilograms of carbon monoxide would be required to reduce the oxide? How many kilograms of coke would be needed? In each case, how many kilograms of pure iron would be produced? First, let's get the symbols out of the way:

Iron(III) oxide: `Fe_(2)O_(3) `

Carbon monoxide: `CO`

And write the balanced equation for the reaction:

`Fe_(2)O_(3)+ 3CO -> 3CO_(2) + 2Fe`

formula weights will be approximately 160 for iron(III)oxide, 28 for carbon monoxide, 44 for carbon dioxide and 53 for iron.  "Exact" formula...

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First, let's get the symbols out of the way:

Iron(III) oxide: `Fe_(2)O_(3) `

Carbon monoxide: `CO`

And write the balanced equation for the reaction:

`Fe_(2)O_(3)+ 3CO -> 3CO_(2) + 2Fe`

formula weights will be approximately 160 for iron(III)oxide, 28 for carbon monoxide, 44 for carbon dioxide and 53 for iron.  "Exact" formula (molar) weights would be 159.6882, 28.0101, 44.0095 and 55.845.  We'll use these latter weights to calculate  our outcome:

carbon monoxide:  (150/159.6882)x3x(28.0101) = 78.9 kg
(2 significant digits, so 79 kg)

iron: (150/159.6882)x2x(55.845) = 104.9 kg

2 significant digits, so 1.0 x 10E2 kg (writing 100 kg would be incorrect)

with coke, essentially pure carbon, the equation becomes

`2Fe_(2)O_(3)+ 3C -> 3CO_(2) + 4Fe`

17 kg of coke would be required to produce `1.0x10^2 `

kilograms of iron. (Again, 100 kg, but it's inaccurate to write it that way).

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