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Al (s) + Fe2O3 (s) --> Al2O3 (s) + Fe (s)

How many grams of Fe can be produced when 69.2 g of Al is reacted with an excess (unlimited) supply of Fe2O3?

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Related questions

In the thermite reaction, iron (III) oxide is reduced by aluminum to give molten iron, Fe2O3 (s) + 2 Al (s) --> 2 Fe (l) + Al2O3 (s) If you begin with 10.0 g of Fe2O3 and 20.0 g Al, Which reactant is limiting? What mass of Fe can be produced? What mass of the excess reactant remains after the limiting reactant is consumed? Set up an amounts table for this problem.

The reaction of iron(III) oxide with aluminum to give molten iron is known as the thermite reaction.

Fe2O3 (s) + 2Al(s) → 2Fe(l) + Al2O3 (s)

What amount of Al, in moles, is needed for complete reaction with 3.0 mol of Fe2O3? What mass of Fe, in grams, can be produced?

If 2.5g of Al (27.0g/mol) and 7.2g of Fe2O3 (159.8g/mol) areplaced in a reaction vesselunder proper conditions, how many gramsof Fe (55.9g/mol) can be produced if the yeild is 100%? How much ofthe reaction in excess is left over?


Balanced Equation:

2Al + Fe2O3 > Al2O3 + 2Fe

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