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 given below.

Fe2O3(s) + 2 Al(s) -->2 Fe(l) + Al2O3(s)

What masses of iron(III) oxide and aluminum must be used to produce 30.0 g iron? What is the maximum mass of aluminum oxide that could be produced?

OK, I'm stuck on the limiting reagent part...

My work:
30g Fe x (1 mol/55.85g Fe) =0.5372

ALUMINUM:
0.5372 mol Fe x (2 mol Al/ 2 mol Fe) = 0.5372 mol Al

0.5372 mol Al x (26.98 g Al/1 mol)=14.194 g Al

IRON(III) OXIDE
0.5372 mol Fe x (1 mol Fe2O3/2 mol Fe)=.2686mol Fe2O3

.2686 mol Fe2O3 x (55.85*2+16*3) g / 1 mol
=42.895 g Fe2O3

What is the maximum mass of aluminum oxide that could be produced? ---limiting reagent problem?

I take all the masses and convert them to moles and divide by their coefficients
30.0 g Fe = .5372 mol Fe / 2=.2686
14.194 g Al = .5261 mol Al / 2 = .2630
42.895 g Fe2O3 = .2686 / 1 = .2686

Their numbers are pretty similar, probably the only difference is the aluminum, so is that the limiting one?

.5261 mol Al x (1 mol Al2O3/2 mol Al) = .2630 mol Al2O3??

.2630 g Al2O3 x (1 mol/101.96 g)= .002579 g

please help me on this last part.

This is not a limiting reagent problem.

The only error I see right off is mass Al needed is closer to 14.5 than to 14.2. I think you just punched the wrong button. As for the maximum amount of Al2O3, just take the moles Fe, convert to moles Al2O3, then to grams and you have it.

thanks!

To determine the limiting reagent, you need to compare the moles of each reactant to the stoichiometry of the balanced equation. From your calculations, you have found that:

- 0.5372 mol Fe
- 0.5261 mol Al
- 0.2686 mol Fe2O3

To find the limiting reagent, compare the mole ratios from the balanced equation:

Fe2O3 : Al = 1 : 2

Since the mole ratio requires 2 moles of Al for every 1 mole of Fe2O3, we can see that we have excess Al because we have 0.5261 mol Al compared to 0.2686 mol Fe2O3. Therefore, Fe2O3 is the limiting reagent.

Now, let's calculate the amounts of Fe2O3 and Al required to produce 30.0 g of iron:

Fe:
- 30.0 g Fe x (1 mol Fe / 55.85 g Fe) = 0.5372 mol Fe

Al:
- 0.5372 mol Fe x (2 mol Al / 2 mol Fe) = 0.5372 mol Al
- 0.5372 mol Al x (26.98 g Al / 1 mol) = 14.494 g Al

So, to produce 30.0 g of iron, you would need approximately 0.2686 mol Fe2O3 and 14.494 g of Al (to ensure you have an excess).

Finally, let's calculate the maximum mass of aluminum oxide that could be produced:

Al2O3:
Since the mole ratio of Al2O3 : Al is 1 : 2, we can calculate it as follows:
- 0.5372 mol Al x (1 mol Al2O3 / 2 mol Al) = 0.2686 mol Al2O3
- 0.2686 mol Al2O3 x (101.96 g Al2O3 / 1 mol) ≈ 27.36 g Al2O3

Therefore, the maximum mass of aluminum oxide that could be produced is approximately 27.36 grams.

To determine the limiting reagent and the maximum mass of aluminum oxide produced, you need to compare the number of moles of aluminum and iron(III) oxide available in the reaction.

From your calculations, you found that you have 0.5261 mol of aluminum and 0.2686 mol of iron(III) oxide.

To determine the limiting reagent, compare the mole ratio of aluminum to iron(III) oxide in the balanced equation. The coefficient for aluminum is 2, while the coefficient for iron(III) oxide is 1.

Divide the moles of aluminum by its coefficient:
0.5261 mol Al / 2 = 0.2631 mol Al

Divide the moles of iron(III) oxide by its coefficient:
0.2686 mol Fe2O3 / 1 = 0.2686 mol Fe2O3

Now, compare these values. The smallest value (0.2631 mol Al) indicates that aluminum is the limiting reagent.

To calculate the maximum mass of aluminum oxide, use the limiting reagent and its coefficient in the balanced equation. The coefficient for aluminum oxide is 1.

Multiply the moles of aluminum oxide by its molar mass:
0.2631 mol Al2O3 x 101.96 g/mol = 26.83 g Al2O3

Therefore, the maximum mass of aluminum oxide that could be produced is 26.83 grams.