# 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)?

Feb 21, 2016

$13.5 g$

#### Explanation:

${n}_{F {e}_{2} {O}_{3}} = \frac{m}{{m}_{r}} = \frac{40}{2 \times 55.8 + 3 \times 16} = 0.25 m o l$.

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, $0.25 m o l$ iron(III)oxide requires $0.50$ moles aluminium.
This corresponds to a mass of :

${m}_{A l} = n \times {M}_{r} = 0.50 \times 27 = 13.5 g$.