One reaction is
Fe2O3 + 3CO = 2Fe + 3CO2
So, figure the number of moles of iron produced, and you will need 1/2 that many moles of Fe2O3.
Then convert that back to mass
Fe2O3 + 3CO = 2Fe + 3CO2
So, figure the number of moles of iron produced, and you will need 1/2 that many moles of Fe2O3.
Then convert that back to mass
The balanced equation for the reduction of iron(III) oxide is:
2 Fe2O3 + 3 C → 4 Fe + 3 CO2
This equation tells us that for every 2 moles of iron(III) oxide, we obtain 4 moles of iron.
Now, let's calculate the molar masses of the compounds involved.
The molar mass of iron(III) oxide (Fe2O3) can be calculated by adding up the atomic masses of its constituent elements, which are:
2 Fe (iron) atoms with a molar mass of 55.845 g/mol (each) = 111.69 g/mol
3 O (oxygen) atoms with a molar mass of 16.00 g/mol (each) = 48.00 g/mol
Total molar mass of Fe2O3 = 111.69 g/mol + 48.00 g/mol = 159.69 g/mol
Next, we need to convert the desired mass from tonnes to grams, using the conversion factor 1 tonne = 1000 kg = 1,000,000 g:
100 tonnes x 1,000,000 g/tonne = 100,000,000 g
Now, let's calculate the number of moles of iron required:
Number of moles of Fe = (mass of Fe / molar mass of Fe) = (100,000,000 g / 55.845 g/mol) = 1,789,188.07 mol
Since the stoichiometric ratio is 2 moles of Fe2O3 to 4 moles of Fe, the number of moles of Fe2O3 needed is half the number of moles of iron:
Number of moles of Fe2O3 = (1/2) x 1,789,188.07 mol = 894,594.04 mol
Finally, let's convert the number of moles of Fe2O3 to grams:
Mass of Fe2O3 = (number of moles of Fe2O3 x molar mass of Fe2O3) = (894,594.04 mol x 159.69 g/mol) = 142,925,828.12 g
Therefore, approximately 142,925,828.12 grams (or about 142.93 tonnes) of iron(III) oxide are needed to produce 100 tonnes of iron.