Iron can be produced from a mixture of aluminum and Fe2O3 according to the following equation.?

Method 1 :

Molar mass of Fe₂O₃ = (55.8×2 + 16.0×3) g/mol = 159.6 g/mol
Molar mass of Al = 27.0 g/mol
Molar mass of Fe = 55.8 g/mol

Initial number of moles of Fe₂O₃ = (85.0 g) / (159.6 g/mol) = 0.533 mol
Initial number of moles of Al = (25.0 g) / (27.0 g/mol) = 0.926 mol

Balanced equation for the reaction :
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
Mole ratio Fe₂O₃ : Al : Fe = 1 : 2 : 2

If 0.926 mol Al completely reacts, Fe₂O₃ needed = (0.926 mol) × (1/2) = 0.463 mol < 0.533 mol
Fe₂O₃ is in excess, and thus Al is the limiting reagent/reactant.

Number of moles of Al reacted = 0.926 mol
Number of moles of Fe produced = 0.926 mol
Mass of Fe produced = (0.926 mol) × (55.8 g/mol) = 51.7 g


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Method 2 :

Molar mass of Fe₂O₃ = (55.8×2 + 16.0×3) g/mol = 159.6 g/mol
Molar mass of Al = 27.0 g/mol
Molar mass of Fe = 55.8 g/mol

Initial number of moles of Fe₂O₃ = (85.0 g) / (159.6 g/mol) = 0.533 mol
Initial number of moles of Al = (25.0 g) / (27.0 g/mol) = 0.926 mol

Balanced equation for the reaction :
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
Mole ratio Fe₂O₃ : Al : Fe = 1 : 2 : 2

If 0.533 mol Fe₂O₃ completely reacts, Fe produced = (0.533 mol) × (2/1) = 1.07 mol
If 0.926 mol Al completely reacts, Fe produced = (0.926 mol) × (2/2) = 0.926 mol < 1.07 mol
Hence, Al is the limiting reactant/reagent.

Number of moles of Fe produced = 0.926 mol
Mass of Fe produced = (0.926 mol) × (55.8 g/mol) = 51.7 g

Thermite reaction.....

The reaction between solid iron(III) oxide and solid aluminum (at least initially) is called a thermite reaction (there are others). It generates a large amount of heat and produces molten iron. Your question is a limiting reactant question. One way to solve it is to compute the amount of product using each of the reactants. The theoretical yield will be the lesser of the two, which also tells you the limiting reactant.

2Al(s) + Fe2O3(s) --> 2Fe(s) + Al2O3(s)
25.0g ... 85.0g .............?g

25.0g Al x (1 mol Al / 27.0g Al) x (1 mol Fe / 1 mol Al) x (55.8g Fe / 1 mol Fe) = 51.7g Fe
85.0g Fe2O3 x (1 mol Fe2O3 / 159.6g Fe2O3) x (2 mol Fe / 1 mol Fe2O3) x (55.8g Fe / 1 mol Fe) = 59.4g Fe

51.7g Fe is the theoretical yield and aluminum is the limiting reactant.

x/[2(55.85)] = 25/[2(27)]
x = 51.71 g <<< round as needed