Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is the following.

Cr2O3(s) + 2 Al(s) → 2 Cr(l) + Al2O3(s)
What masses of chromium(III) oxide and aluminum must be used to produce 15.4 g of chromium?
What is the maximum mass of aluminum oxide that could be produced?

To calculate the masses of chromium(III) oxide and aluminum needed to produce 15.4 g of chromium, and to determine the maximum mass of aluminum oxide that could be produced, we can use stoichiometry.

1. Calculate the molar masses:
- Chromium(III) oxide (Cr2O3) = molar mass of Cr + 3 * molar mass of O = (52.00 g/mol) + (16.00 g/mol) * 3 = 152.00 g/mol
- Aluminum (Al) = 26.98 g/mol

2. Determine the stoichiometric ratio between chromium and chromium(III) oxide:
From the balanced equation, we can see that 1 mol of chromium(III) oxide produces 2 mol of chromium.

3. Calculate the amount of chromium(III) oxide required:
Mass of chromium(III) oxide = (15.4 g chromium) * (1 mol Cr2O3 / 2 mol Cr) * (152.00 g/mol) = 235.60 g

Therefore, approximately 235.60 grams of chromium(III) oxide should be used to produce 15.4 grams of chromium.

4. Determine the stoichiometric ratio between aluminum and aluminum oxide:
From the balanced equation, we can see that 2 mol of aluminum produces 1 mol of aluminum oxide.

5. Calculate the maximum amount of aluminum oxide produced:
Mass of aluminum oxide = (mass of aluminum used) * (1 mol Al2O3 / 2 mol Al) * (101.96 g/mol)
To find the maximum mass, we need to determine the limiting reagent.

To do this, we'll calculate the number of moles of both reactants (aluminum and chromium(III) oxide), and compare them based on their stoichiometric ratios.

Number of moles of aluminum = (mass of aluminum used) / (molar mass of aluminum)
Number of moles of chromium(III) oxide = (mass of chromium(III) oxide used) / (molar mass of chromium(III) oxide)

Compare the ratios:
Moles of aluminum / Stoichiometric coefficient of aluminum = Moles of chromium(III) oxide / Stoichiometric coefficient of chromium(III) oxide

Compare:
(Number of moles of aluminum) / 2 = (Number of moles of chromium(III) oxide) / 1

From here, solve for the mass of aluminum:
Mass of aluminum = [(Number of moles of aluminum) / 2] * (molar mass of aluminum)

By comparing the mass of aluminum obtained from this calculation with the amount of aluminum used, we can determine if aluminum is in excess or if chromium(III) oxide is the limiting reagent.

6. Calculate the maximum mass of aluminum oxide:
Mass of aluminum oxide = (mass of aluminum used) * (1 mol Al2O3 / 2 mol Al) * (101.96 g/mol)

Therefore, the maximum mass of aluminum oxide that could be produced can be determined by using the mass of aluminum obtained from the previous calculation.