# How does aluminum react with ferric oxide?

Dec 24, 2016

$F {e}_{2} {O}_{3} + 2 A l \rightarrow 2 F e + A {l}_{2} {O}_{3} + \Delta$
$\text{Moles of aluminum}$ $=$ $\frac{40.5 \cdot g}{27.0 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.5 \cdot m o l$
And, clearly, given the stoichiometry, $0.75 \cdot m o l$ of ferric oxide are required for equivalence, i.e. $0.75 \cdot m o l \times 159.69 \cdot g \cdot m o {l}^{-} 1$ $\cong 120 \cdot g$.