Question

How many grams of Al2O3 can form when 23.6 g Al react with excess Fe2O3?

I am so lost...please show your work with a drawing

Answer 1

First let's come up with the balanced chemical reaction we're dealing with. What do you think it is?

Answer 2

Fe2O3 + 2 Al --> 2 Fe + Al2O3, Next I have to balance it,

Answer 3

Oh derp, it is already balanced.

Answer 4

What do I do next?

Answer 5

We want to find out how much Al2O3 is produced. That means we need to know how much Al reacted. We can find this using the chemical equation you've given. However, keep in mind the chemical reaction uses numbers of molecules, not masses. That means we need to convert the mass of Al to a number, specifically moles.

How many moles in 23.6 g of Al?

Answer 6

One mole of Al is equal to 26.981g, since that is the atomic mass, so, 23.6/26.981 is equal to 0.874

Answer 7

*.875

Answer 8

Ok so we have 0.875 mol of Al. How much Al2O3 can be produced from this? HINT: From your balanced equation, you can see that for every 2 moles of Al that react, only 1 mole of Al2O3 is produced.

Answer 9

0.475 moles of Al2O3?

Answer 10

*0.4375, I think you just dropped the 3.

So that's moles of Al2O3, and now we just need the mass. How can you go from moles back to mass?

Answer 11

So that is 0.4375 moles of Al2O3, now I need to convert that back to mass

Answer 12

I need to calculate the molar mass of the entire substance correct?

Answer 13

So the grams of each element separtely

Answer 14

so, 26.981 times 2(Al2) is 53.962

Answer 15

So 53.962+the mass of O3, which is approximately 16 grams per Oxygen atom, 16 times 3 is 48, so 53.96+48

Answer 16

That is 101.96. This is 1 mole of Al2O3, I do not have 1 mole though so, I have to turn that into 0.4375 moles.

Answer 17

So, that would be with multiplication, so 101.96 times 0.4375 would be 44.6075

Answer 18

Exactly right!