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

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To solve the reaction Fe2O3 + 2Al → 2Fe + Al2O3 with 3.3 mol of Fe2O3, 6.6 moles of aluminum are needed due to the stoichiometric ratio. This also results in the formation of 6.6 moles of iron. The moles of aluminum oxide produced will be 3.3, matching the initial amount of Fe2O3. The reasoning for aluminum oxide is clarified as it corresponds to the coefficient in the balanced equation rather than the total moles used. Understanding the stoichiometry of the reaction is crucial for accurate calculations.
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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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