Chemistry - yields question

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)

Reply 2

OP

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)

Reply 3

OP

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!

Reply 4

OP

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...

Reply 5

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)

Reply 6

OP

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...

Reply 7

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.

Reply 8

OP

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?

Reply 9