Question

How is the oxidation of aluminum metal represented?

Answer 1

`4Al+3O_">2Al_2O_3`

Explanation:

At first we have `Al+O_2>Al_2O_3`

Let's first balance `Al`, the LHS has 1 Al atom, while the RHS has 2. To balance this we add another `Al` to the LHS, to get: `2Al+O_2>Al_2O_3`

Now to balance `O`, the LHS has 2, while the RHS has 3, `2*3=6`, so we have `2Al+3O_2>2Al_2O_3`

Mow `Al` is unbalanced again, to balance, we just add an extra 2 `Al` to the LHS to get `4Al + 3O_2>2Al_2O_3`.

Answer 2

`"Garbage in must equal garbage out......."`

Explanation:

`2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)`

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

`4Al(s) + 3O_2(g) rarr 2Al_2O_3(s)`

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, Cl_2` are BINUCLEAR.