Al + O2 → Al2O3 limiting reagents?

Which is the limiting reactant when 0.32 mol Al and 0.26 mol O2 are available?

How many moles of Al2O3 are formed from the reaction of 6.38 mol O2 and 9.15 mol of Al?


I think I know the answers, I just need to know how to work them out. Any help would be appreciated. Thank you in advance.

Answer:
2Al + 3O2 ----> 2Al2O3 (balanced equation)

a. 0.32 mol Al x (2 mol Al2O3 / 2 mol Al) = 0.32 mol Al
0.26 mol O2 x (2 mol Al2O3 / 3 mol O2) = 0.17 mol O2
limiting reagent: O2

b. 6.38 mol O2 x (2 mol Al2O3 / 3 mol O2) = 4.25 mol Al2O3
9.15 mol Al x (2 mol Al2O3 / 2 mol Al) = 9.15 mol Al2O3
O2 is the limiting reagent...
4.25 mol Al2O3 is formed from the reaction of 6.38 mol O2 and 9.15 mol of Al.
the one with less no. of moles is the limiting reagent

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