Question

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)`?

Answer 1

`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`.