# How many atoms of carbon are needed to produce 0.45 mol aluminium?

How many atoms of carbon are needed to produce $\pu{0.45 mol} ~\ce{Al}$? $$\ce{3C + 2Al2O3 -> 4Al + 3CO2}$$

I am not sure where to start, whether or not I convert moles straight to atoms or convert moles to grams of $\ce{Al}$, then to moles of $\ce{C}$ and then to atoms? I tried the latter of my suggestions and ended up with $\pu{9.8E24}$, but I don't know if this is correct.

• Stoichometry is done by comparing numbers of things to one another. We could do stoichiometry with number of atoms or with moles (remember that moles are really just grouping numbers like dozens.) We cannot convert among species using grams though, since all of the different reactants and products have different masses per particle. – Jason Patterson Sep 24 '14 at 16:10

The key to this is by examining the mole ratio. Your balanced equation shows that 4 mols aluminium: 3 mols carbon

If you are familiar with ratios from maths, you can therefore find the specific number of moles of carbon in this reaction.

$4\ce{Al}: 3\ce{C}\implies0.45\ce{Al} :~?$

I think you should be able to get the answer as 0.3375 moles of carbon (again, it's just ratios)! From here, you use moles = mass / mr to get your answer.

You can't reduce aluminium oxide to aluminium using carbon; aluminium, like many metals, forms a stable carbide. The equation looks like this: $\ce{2 Al2O3 + 9 C \rightarrow Al4C3 + 6 CO}$

That's why in real life aluminium is prepared through the Hall–Héroult process. Before the electrochemical process was discovered the reducing agent used was sodium.

• I downvoted this answer because, while this may indeed be true, I am not sure how this answers a high school stoichiometry question, in which we have to base our answer on another given reaction. This answer of yours is better fits as a comment on the original post. Thank you for your cooperation! – Gaurang Tandon Feb 27 '18 at 8:15