# Question cec43

Oct 24, 2017

$= 27.98 g F e$

#### Explanation:

Rewrite your compound so you can visually see how many moles you have of each element.

Iron (III) Oxide $\to F {e}_{2} {O}_{3}$

Then use basic conversion factors:

Iron has 55.845 g/(mol.
Oxygen has 15.99 g/(mol#.

There are 2 moles of Iron and 3 moles of Oxygen in Iron (III) Oxide.

We want to end up with grams of iron.

$2 \left(55.845\right) + 3 \left(15.99\right) = 159.66$

$159.66 g$ $F {e}_{2} {O}_{3}$ per mole.

$40.0 g$ $F {e}_{2} {O}_{3} \cdot \frac{1 m o l F {e}_{2} {O}_{3}}{159.66 g F {e}_{2} {O}_{3}} \cdot \frac{2 m o l F e}{1 m o l F {e}_{2} {O}_{3}} \cdot \frac{55.845 g F e}{1 m o l F e}$

$= 27.98 g F e$