Mass of an element required for a chemical reaction equation

[Kiah Palmer]

Homework Statement

Calculate the mass of 80% pure aluminium required to produce 9.65 kg of aluminium oxide. Hint: First calculate the mass of pure aluminium.

Relevant Equations

Al + O2 -> Al2O3 = 4Al + 3O2 -> 2Al2O3

Conversion: 9.65 kg = 9650 g

Mass Al = 26.98 g/mol

Molar Mass Al2O3 = (2 x 26.98) + ( 3 x 16.00) = 101.96 g/mol

Mol Al2O3 = 9650 g / 101.96 g/mol = 94.65 mol

Mol Al Required = 94.65 mol x 2 = 189.3 mol

Mass 100% Al required = 189.3 mol x 26.98 g/mol = 5.107x103 g

Mass 80% Al required = 5.107x103 g x (80/100) = 4.086x103

I believe this is correct, however, if it isn't could someone please let me know where I went wrong. I will be handing it in as is, but it would still be nice to know. I had looked up the answer after completing it, but the way that person completed it doesn't look right to me ( and has a different answer for the last section. I will include the answer I had found below, and you will notice that I had most of it, however, if their last section is correct then an explanation of how would be greatly appreciated. Thanks!The molar mass of Al2O3 is 101.96 g/mole
Mass of aluminium = 26.98 g/mole.
So 9650 g Al2O3 = 94,645 moles.
And to make 1 mole Al2O3 you need 2 moles Al.
Then you need 189,29 moles Al = 189,29*26,98 g = 5107,04 g.

And if 5107,04 g are 80%, then is 5107,04*100/80 = 100% = 6383,8 g needed.

 

[Borek]

Al + O2 -> Al2O3 = 4Al + 3O2 -> 2Al2O3

OK

Mass 100% Al required = 189.3 mol x 26.98 g/mol = 5.107x103 g

10³, please format properly (but yes, the result is OK)

Mass 80% Al required = 5.107x103 g x (80/100) = 4.086x103

That's 80% of a required mass, and you need something 80% of which is the required mass, do you see the difference?

 

[Kiah Palmer]

Ah! ok, so the answer that I had found when looking afterwards is in fact the correct answer! It actually makes sense now! thanks again! :) I did hand it in the way I Have done it, but at least now I will know where I went wrong. :)