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.

Question

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?

Answer 1

First, write the balanced equations:

Fe₂O₃ + 3CO --> 2Fe + 3CO2

2Fe₂O₃ + 3C --> 4Fe + 3CO₂

Next, set up the calculations:

x kg_CO = 150 kg_Fe2O3 ( 1 mol_Fe2O3 / .1596882 kg_Fe2O3 ) ( 3 mol_CO / 1 mol_Fe2O3 ) ( .0280101 kg_CO / mol_CO ) = 78.9 kg_CO

 

x kg_C = 150 kg_Fe2O3 ( 1 mol_Fe2O3 / .1596882 kg_Fe2O3 ) ( 3 mol_C / 2 mol_Fe2O3 ) ( .0120107 kg_C / mol_C ) = 16.9 kg_C

 

To find the mass of iron formed in each case, make note that 2 moles of iron are formed from each mole of iron (III) oxide:

x kg_Fe = 150 kg_Fe2O3 ( 1 mol_Fe2O3 / .1596882 kg_Fe2O3 ) ( 2 mol_Fe / 1 mol_Fe2O3 ) ( .055845 kg_Fe / mol_Fe ) = 105 kg_Fe