# Chemistry - yields question

Aluminium is produced by the electrolysis of aluminium oxide. The equation for
the reaction is:
2Al2O3 +4Al --> 3O2
Calculate the percentage yield for the reaction if 1078 g of aluminium oxide
produced 539 g of aluminium

How do I work out the moles first, I know the steps for a calculation like this but do I have to include the large numbers used to balance the equation or not?
Original post by murrayma767
Aluminium is produced by the electrolysis of aluminium oxide. The equation for
the reaction is:
2Al2O3 +4Al --> 3O2
Calculate the percentage yield for the reaction if 1078 g of aluminium oxide
produced 539 g of aluminium

How do I work out the moles first, I know the steps for a calculation like this but do I have to include the large numbers used to balance the equation or not?

Start by calculating the moles of aluminium oxide used.

EDIT: I presume the equation is meant to read as

2Al2O3 —> 4Al + 3O2
(edited 6 months ago)
moles of Al2O3 = 1078/2(27) + 3(16)
= 10.56827...
Is that right or should it be:
moles of Al2O3 = 1078/2(2(27) + 3(16))

= 5.284...
(edited 6 months ago)
Original post by TypicalNerd
Start by calculating the moles of aluminium oxide used.

EDIT: I presume the equation is meant to read as

2Al2O3 —> 4Al + 3O2

Yes, sorry!
Original post by murrayma767
moles of Al2O3 = 1078/2(27) + 3(16)
= 10.56827...
Is that right or should it be:
moles of Al2O3 = 1078/2(2(27) + 3(16))

= 5.284...

I have tried continuing with my method but neither of these solutions work...
Original post by murrayma767
Yes, sorry!

So have you worked out the moles of Al2O3 yet? This is important for working out the theoretical yield

Edit: my TSR is being slow, I can see you have given it a shot
(edited 6 months ago)
Original post by TypicalNerd
So have you worked out the moles of Al2O3 yet? This is important for working out the theoretical yield

are either of these solutions correct?
moles of Al2O3 = 1078/2(27) + 3(16)
= 10.56827...
Is that right or should it be:
moles of Al2O3 = 1078/2(2(27) + 3(16))

= 5.284...
Original post by murrayma767
are either of these solutions correct?
moles of Al2O3 = 1078/2(27) + 3(16)
= 10.56827...
Is that right or should it be:
moles of Al2O3 = 1078/2(2(27) + 3(16))

= 5.284...

You’ve correctly worked out the moles of Al2O3

To get the theoretical yield of Al, you double the moles of Al2O3, because the equation shows that for every 2 mol of Al2O3 that react, you make 4 mol of Al. Instead, you appear to have halved it.
OK so:
moles of Al2O3 = 10.56827...
moles of Al = 10.56827x2 = 21.13654
theoretical mass of Al = moles x mr
= 21.13654 x 27 = 570.68658 g
but this is more than the yield given in the question?
Original post by murrayma767
OK so:
moles of Al2O3 = 10.56827...
moles of Al = 10.56827x2 = 21.13654
theoretical mass of Al = moles x mr
= 21.13654 x 27 = 570.68658 g
but this is more than the yield given in the question?

That is perfectly correct. The theoretical yield is almost always greater than the actual yield as it represents the absolute most amount of product you could possibly get.

539 grams is what percentage of 570.68658 g?
I got 94.44763... %
= 94.4 %
Original post by murrayma767
I got 94.44763... %
= 94.4 %

I believe that is correct. There’s your percentage yield.
Thank you so much for your help