# 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...

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?

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### 2 Answers

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).

The balanced reaction of Iron(III) Oxide with Carbon monoxide is

`Fe_2O_3 + 3COrArr 2Fe + 3CO_2`

1 Mole of Iron(III) Oxide require 3 Moles of Carbon monoxide to produce 2 moles of Iron.

The molecular mass of Iron(III) Oxide is 159.69 g/mol , Molecular mass of Carbon monoxide is 28.01 g/mol and Atomic mass of Iron is 55.845.

159.69 g of Iron(III) Oxide requires 3*28.01 g of Carbon monoxide

150 kg of Iron(III) oxide will require `((3*28.01)/159.69)*150` kg of Carbon monoxide

**78.93 kg of Carbon monoxide will be required**

1 Mole of Iron(III) Oxide will produce 2 moles of Iron

159.69 g of Iron(III) Oxide will produce 2*55.845 g of Iron

150 kg of Iron(III) Oxide will produce (2*55.845*150)/159.69 kg of Iron

**150 kg of Iron(III) Oxide will produce 104.91 kg of Iron**

The balanced reaction of Iron(III) Oxide with Carbon is

`2Fe_2O_3 +3C-> 4Fe+ 3CO_2`

2 Moles of Iron(III) Oxide require 3 Moles of Carbon to produce 4 Moles of Iron

2*159.69 g of Iron(III) Oxide requires 3*12.01 g of Carbon

150 kg of Iron(III) Oxide will require (3*12.01*150)/(2*159.69) kg of Carbon

**150 kg of Iron(III) oxide will require 16.92 kg of Carbon**

**Iron produced will be same as in first case since 1 mole of Iron (III) Oxide produces 2 moles of Iron . Iron produced will be 104.91 kg**