Aluminum metal and Iron(III) oxide react to give aluminum and iron metal. What is the maximum amount of Iron that can be obtained from a reaction mixture containing 2.5 grams of aluminum and 9.5 grams of Iron(III) oxide? What mass of excess reactant remains?

You made a typo. It should read, Aluminum metal and iron(III) oxide...........give aluminum OXIDE and iron metal.....

I work these limiting reagent problems by solving TWO stoichiometry problems. First use grams Al and convert to moles iron assuming you had all of the iron oxide you need. Then use grams iron oxide assume you have all of the aluminum you need and convert to moles Fe. The answers for mols Fe will be different but the correct answer in limiting reagent problems is ALWAYS the smaller value and the substance providing that value is the limiting reagent. The other reagent is the one in excess.
Here is a worked example of the simple stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Post your work if you get stuck.

To determine the maximum amount of iron that can be obtained from the reaction, we need to first identify the limiting reactant. The limiting reactant is the reactant that is completely consumed, thereby limiting the amount of product that can be formed. The reactant that is not fully consumed is called the excess reactant.

Let's start by calculating the number of moles for each reactant using their respective molar masses.

1. Aluminum (Al):
The molar mass of aluminum is 26.98 g/mol.
Number of moles of aluminum = mass of aluminum / molar mass of aluminum
Number of moles of aluminum = 2.5 g / 26.98 g/mol

2. Iron(III) oxide (Fe2O3):
The molar mass of Fe2O3 can be calculated by adding the molar masses of iron (Fe) and oxygen (O).
Molar mass of Fe2O3 = (2 × molar mass of Fe) + (3 × molar mass of O)
Molar mass of Fe2O3 = (2 × 55.85 g/mol) + (3 × 16.00 g/mol) = 159.69 g/mol
Number of moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
Number of moles of Fe2O3 = 9.5 g / 159.69 g/mol

Now, let's determine the stoichiometry of the balanced chemical equation to find the mole ratio between aluminum and iron. The balanced equation is:

2Al + Fe2O3 → 2Fe + Al2O3

According to the equation, the mole ratio between aluminum and iron is 2:2, which simplifies to 1:1.

Comparing the number of moles of aluminum and Fe2O3, we can see that the mole ratio is 1:2. This means that for every 1 mole of aluminum, we need 2 moles of Fe2O3 to react completely.

Since the mole ratio is 1:2, this indicates that aluminum is the limiting reactant since we have fewer moles of aluminum compared to Fe2O3.

To determine the maximum amount of iron that can be obtained, we need to consider the stoichiometry of the balanced equation. Since the mole ratio between aluminum and iron is 1:1, the number of moles of iron formed will be equal to the number of moles of aluminum.

Using the number of moles of aluminum, we can calculate the mass of iron formed by multiplying it by the molar mass of iron (55.85 g/mol).

Mass of iron formed = Number of moles of aluminum × molar mass of iron
Mass of iron formed = (2.5 g / 26.98 g/mol) × 55.85 g/mol

To find the mass of excess reactant remaining, we can subtract the mass of iron formed from the mass of iron(III) oxide initially used.

Mass of excess reactant remaining = Mass of iron(III) oxide - Mass of iron formed
Mass of excess reactant remaining = 9.5 g - (2.5 g / 26.98 g/mol) × 55.85 g/mol

Calculating the above equations will give you the maximum amount of iron that can be obtained from the reaction and the mass of excess reactant remaining.