What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed in the reaction #Fe_2O_3(s) + 2Al(s) -> 2Fe(l) + Al_2O_3(s)#?

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What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed in the reaction #Fe_2O_3(s) + 2Al(s) -> 2Fe(l) + Al_2O_3(s)#?

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Answer 1 · 2016-02-21T19:05:38

#13.5g#

Explanation:

#n_(Fe_2O_3)=m/(m_r)=40/(2xx55.8+3xx16)=0.25mol#.

The balanced chemical equation represents the mole ratio in which chemicals combine.

Hence 1 mole iron(III)oxide requires 2 moles aluminium.
Therefore by ratio and proportion, #0.25 mol# iron(III)oxide requires #0.50# moles aluminium.
This corresponds to a mass of :

#m_(Al)=nxxM_r=0.50xx27=13.5g#.

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What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed in the reaction #Fe_2O_3(s) + 2Al(s) -> 2Fe(l) + Al_2O_3(s)#?

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