# Using the equation, Fe_2O_3 + 2Al -> 2Fe + Al_2O_3, how many grams of Al are needed to completely react with 135 grams of Fe_2O_3?

$135 g F {e}_{2} {O}_{3} \cdot \frac{1 m o l F {e}_{2} {O}_{3}}{159.69 g F {e}_{2} {O}_{3}} \cdot \frac{2 m o l A l}{1 m o l F {e}_{2} {O}_{3}} \cdot \frac{26.98 g A l}{1 m o l A l} = 45.6 g A l$
Then, by looking at the relationship between the moles of iron (III) oxide and aluminum, you can determine the number of moles of aluminum per mole of $F {e}_{2} {O}_{3}$.
Finally, convert the number of moles of aluminum to grams of aluminum by multiplying by its molar mass. After unit cancellation, you can determine that 45.6 g Al is necessary to completely react with 135 g $F {e}_{2} {O}_{3}$.