The following redox reaction is highly exothermic and is used to produce molten iron for welding pieces of steel together, e.g - Leaving Cert Chemistry - Question b - 2013

In the given reaction:

• Aluminum (Al) changes from an oxidation state of 0 in elemental form to +3 in aluminum oxide (Al₂O₃), thus it is oxidized.
• Iron (Fe) changes from an oxidation state of +3 in iron(III) oxide (Fe₂O₃) to 0 in elemental iron (Fe), thus it is reduced. This indicates that there is a transfer of electrons, confirming that this is a redox reaction.

The reducing agent in this reaction is aluminum (Al) because it donates electrons and is oxidized in the process.

To calculate the mass of aluminum required to produce 1008 g of iron:

1. From the balanced equation, 8 mol of Al produces 9 mol of Fe.
2. Calculate the molar mass of Fe:
   • Molar mass of Fe = 56 g/mol
   • 9 mol Fe = 9 × 56 g = 504 g.
3. If 8 mol of Al yields 504 g of Fe, the amount of Al needed for 1008 g of Fe: \text{Mass of Al} = \left( #\frac{8 \text{ mol Al}}{9 \text{ mol Fe}} \right) \times \frac{1008 \text{ g Fe}}{56 \text{ g/mol}} $= \left( \frac{8}{9} \right) \times 18 \text{ mol Al} = 16 \text{ mol Al}$
4. Molar mass of Al is 27 g/mol, therefore: $\text{Mass of Al} = 16 \text{ mol} \times 27 \text{ g/mol} = 432 ext{ g}$

To find the mass of aluminum oxide (Al₂O₃) produced:

1. From the balanced equation, 8 mol of Al produces 4 mol of Al₂O₃.
2. Since 16 mol of Al is used, the moles of Al₂O₃ produced: $\text{Moles of Al}_2O_3 = \left( \frac{4 \text{ mol Al}_2O_3}{8 \text{ mol Al}} \right) \times 16 \text{ mol Al} = 8 \text{ mol Al}_2O_3$
3. The molar mass of Al₂O₃:
   • Molar mass = (2 × 27 g/mol) + (3 × 16 g/mol) = 54 g + 48 g = 102 g/mol.
4. Mass of Al₂O₃ produced: $\text{Mass of Al}_2O_3 = 8 \text{ mol} \times 102 \text{ g/mol} = 816 ext{ g}$