# How many moles of rust (Fe_2O_3) will be formed by the rusting of 25.0 grams of iron (Fe)?

Approx. $\frac{1}{4}$ $m o l$ of ferric oxide.
$2 F e \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow F {e}_{2} {O}_{3} \left(a q\right)$
Moles of iron $=$ $\frac{25.0 \cdot g}{55.85 \cdot g \cdot m o {l}^{-} 1}$ $=$ ?? $\text{mol}$