# How is the oxidation of aluminum metal represented?

Jul 23, 2017

4Al+3O_">2Al_2O_3

#### Explanation:

At first we have $A l + {O}_{2} > A {l}_{2} {O}_{3}$

Let's first balance $A l$, the LHS has 1 Al atom, while the RHS has 2. To balance this we add another $A l$ to the LHS, to get: $2 A l + {O}_{2} > A {l}_{2} {O}_{3}$

Now to balance $O$, the LHS has 2, while the RHS has 3, $2 \cdot 3 = 6$, so we have $2 A l + 3 {O}_{2} > 2 A {l}_{2} {O}_{3}$

Mow $A l$ is unbalanced again, to balance, we just add an extra 2 $A l$ to the LHS to get $4 A l + 3 {O}_{2} > 2 A {l}_{2} {O}_{3}$.

Jul 23, 2017

$\text{Garbage in must equal garbage out.......}$

#### Explanation:

$2 A l \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow A {l}_{2} {O}_{3} \left(s\right)$

If the non-stoichiometric coefficients offend thine sensibilities.....then DOUBLE the entire equation......

$4 A l \left(s\right) + 3 {O}_{2} \left(g\right) \rightarrow 2 A {l}_{2} {O}_{3} \left(s\right)$

In either instance, 2 equiv of aluminum metal are oxidized by 3 equiv dioxygen gas to give one equiv alumina.

Note that with regard to dioxygen gas, ALL of the elemental gases (save for the Noble Gases) are diatomic, e.g. ${H}_{2} , {N}_{2} , {O}_{2} , {F}_{2} , C {l}_{2}$ are BINUCLEAR.