How much iron can be prepared from a 1*kg mass of "ferric oxide"?

May 8, 2017

Well, how many moles of $\text{ferric oxide}$ do you have...........?

Explanation:

$\text{Moles of ferric oxide} = \frac{1000 \cdot g}{159.69 \cdot g \cdot m o {l}^{-} 1} = 6.262 \cdot m o l$.

I converted this to $\text{grams}$, because, as chemists, we typically deal with gram quantities. And thus if there are $6.262 \cdot m o l$ of $F {e}_{2} {O}_{3}$, there are $12.52 \cdot m o l$ with respect to iron, and $18.79 \cdot m o l$ with respect to oxygen. Here we use the molar quantity, the NUMBER of iron and oxygen atoms as our yardstick, and of course we use the stoichiometric proportion of atoms in formula.

And thus AT MOST, we can get $12.52 \cdot m o l$ of iron metal upon reduction.

This represents a mass of $12.52 \cdot m o l \times 55.85 \cdot g \cdot m o {l}^{-} 1$, approx. $700 \cdot g$ iron. Are you with me?