The thermite reaction, used to weld rails together in the building of railroads, is described by the following equation.
Fe2O3(s) + 2Al(s) --> Al2O3(s) + 2Fe(l)
Calculate the mass of iron metal that can be prepared from 150 grams of aluminum and 250 grams of iron (III) oxide.
250gFe2O3 x (1moleFe2O3 / 85.9995gFe2O3) x (2moleFe / 1moleFe2O3) x (18.99840gFe) = 110gFe
150gAl x (1moleAl / 26.98154gAl) x (2moleFe / 2mole Al) x (18.99840gFe / 1moleFe) = 53gFe
I know the limiting reactant is Al but I do not know how to get the mass of iron metal that is prepared from the aluminum and iron (III) oxide. I know the answer is supposed to be 175gFe. What should I do to get this? Please explain. Thanks
I don't know what you've done. What's the 18.998? Where did you come up with 85.9995? The molar mass of Fe2O3 is 159.7. The atomic mass Fe is 55.85. How do you know Al is the limiting reagent. I think it is Fe2O3.
If you know Al is the limiting reagent, go from there this way.
mols Fe2O3 = 250/159.7 = ?
mols Fe = 2 x that.
grams Fe = mols Fe x 55.85 = 174.85 grams which rounds to 175g.
Thank you! I think I understand it a bit better
To calculate the mass of iron metal that is prepared from aluminum and iron (III) oxide, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the amount of product that can be formed.
In this case, we have 150 grams of aluminum (Al) and 250 grams of iron (III) oxide (Fe2O3).
1. Convert the masses of aluminum and iron (III) oxide to moles.
150g Al x (1 mol Al / 26.98154g Al) = 5.56 mol Al
250g Fe2O3 x (1 mol Fe2O3 / 159.6882g Fe2O3) = 1.57 mol Fe2O3
2. Use the stoichiometry of the balanced equation to determine the amount of iron (Fe) that can be produced from each reactant.
From the balanced equation: 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l)
For aluminum:
5.56 mol Al x (2 mol Fe / 2 mol Al) = 5.56 mol Fe
For iron (III) oxide:
1.57 mol Fe2O3 x (2 mol Fe / 1 mol Fe2O3) = 3.14 mol Fe
3. Compare the amounts of iron (Fe) obtained from each reactant. The limiting reactant is the one that produces less iron (Fe).
Since we have less iron (Fe) obtained from aluminum (5.56 mol) compared to iron (III) oxide (3.14 mol), aluminum is the limiting reactant.
4. Calculate the mass of iron (Fe) produced from the limiting reactant.
Using the molar mass of iron (Fe) (55.845 g/mol):
5.56 mol Fe x (55.845 g Fe / 1 mol Fe) = 310 g Fe
Therefore, the mass of iron metal produced from aluminum and iron (III) oxide is 310 grams, not 175 grams.
It's possible that there might be a mistake in the calculations you provided, as the correct answer should be 310 grams, not 175 grams. Please double-check the calculations or provide further information if there is any additional context or parameters to consider.