Solving Thermite Reaction: 1 Mole Al + 10.0g Fe2O3

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The discussion focuses on calculating the amount of unreacted aluminum after a thermite reaction involving iron(III) oxide (Fe2O3) and aluminum (Al). Given that 10.0 grams of Fe2O3 corresponds to approximately 0.0626 moles, it is established that this amount of Fe2O3 will react with twice the moles of aluminum, resulting in 0.1252 moles of Al consumed. Starting with 1 mole of Al, the remaining amount after the reaction is calculated to be about 0.8748 moles, which translates to approximately 23.6 grams of unreacted aluminum.
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Please Help!

I am still having problems with this problem...

It deals with the thermite reaction Fe_{2}O_{3}+2Al----2Fe+Al_{2}O_{3}

If 1 mole of Al is reacted with 10.0g of Fe_{2}O_{3}, how many grams of unreacted Al would remain?
 
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How many moles of Fe_2O_3 are there in 10g?
 
There are about 0.0626 moles of Fe_{2}O_{3} in 10.0g
 
This is a double post.
Anyway no:moles of Fe_2O_3 reacted will be twice no:moles of aluminium.
if only 0.0626 moles of Fe_2O_3 is present then correspondingly 0.0626*2 moles of Al will react.
 
Al remaining

Assuming 10 g of Fe2O3 = 0.0626 moles (too lazy to look at periodic table),
then Al remains is (1 - (2*0.0626))= 0.8748 mol of Al = approx. 23.6 g of Al
 
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