Solving Thermite Reaction: 1 Mole Al + 10.0g Fe2O3

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Discussion Overview

The discussion revolves around calculating the amount of unreacted aluminum remaining after a thermite reaction involving 1 mole of aluminum and 10.0 grams of iron(III) oxide (Fe2O3). The focus is on stoichiometry and the application of the reaction equation.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant seeks help with the stoichiometric calculation of unreacted aluminum after the thermite reaction.
  • Another participant questions how many moles of Fe2O3 are present in 10 grams.
  • A subsequent reply estimates that there are about 0.0626 moles of Fe2O3 in 10.0 grams.
  • Another participant clarifies that the moles of Fe2O3 reacted will be half the number of moles of aluminum, based on the reaction stoichiometry.
  • One participant calculates the remaining aluminum, concluding that approximately 23.6 grams of aluminum would remain unreacted.

Areas of Agreement / Disagreement

Participants generally agree on the stoichiometric relationships in the reaction, but there is no consensus on the final amount of unreacted aluminum, as calculations vary slightly.

Contextual Notes

Assumptions include the accuracy of the molar mass used for calculations and the precision of the initial mass of Fe2O3. The discussion does not resolve any potential discrepancies in these calculations.

Who May Find This Useful

Students and individuals interested in stoichiometry, chemical reactions, and thermite reactions may find this discussion relevant.

chemister
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Please Help!

I am still having problems with this problem...

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

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