Solving 3.3 mol Fe_{2}O_{3}: 2Fe+Al_{2}O_{3}

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The discussion focuses on calculating the amounts of reactants and products in the chemical reaction Fe2O3 + 2Al → 2Fe + Al2O3 when starting with 3.3 moles of Fe2O3. The calculations conclude that 6.6 moles of aluminum are required, 6.6 moles of iron are produced, and 3.3 moles of aluminum oxide are formed. The reasoning for the aluminum oxide amount is clarified, emphasizing that it corresponds directly to the initial moles of Fe2O3 due to the stoichiometry of the reaction.

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Calculate the amounts requested if 3.3 mol of Fe_{2}O_{3}
Fe_{2}O_{3}+2Al\rightarrow2Fe+Al_{2}O_{3}

a)Find the moles of aluminum needed
b)Find the moles of iron formed
c)Find the moles of aluminum oxide formed


Not sure how to go about solving any of these the book is terrible and doesn't explain anything and I can't seem to find good notes online.

Im thinking that
a)6.6 moles because there is a 2 infront
b)6.6 again same reason
c)3.3 to even out the moles used
 
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Good job, your answers are exactly correct.
But your reasoning might be a little off on part 'c'. You end up with 3.3 moles of aluminum oxide because it has the same number in front as the Fe2_O3 you started with.

Consider this example:
A + B \rightarrow AB
There are two moles of substance on the left, but only one on the right---and that is totally fine. The number of moles of particular substances doesn't need to match on each side---as long as the number of atoms matches (there is 1 mol of A, and 1 mol of B on each side).
 

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