# What is the mass of iron that could be obtained from 67.0 grams of iron(III) oxide?

Jan 31, 2016

$67.0$ $g$ of iron(III) oxide represents $\frac{67.0 \cdot \cancel{g}}{159.69 \cdot \cancel{g} \cdot m o {l}^{-} 1}$, approx. $0.4$ $m o l$.