How is the oxidation of aluminum metal represented?

2 Answers
Jul 23, 2017

4Al+3O_">2Al_2O_34Al+3O>2Al2O3

Explanation:

At first we have Al+O_2>Al_2O_3Al+O2>Al2O3

Let's first balance AlAl, the LHS has 1 Al atom, while the RHS has 2. To balance this we add another AlAl to the LHS, to get: 2Al+O_2>Al_2O_32Al+O2>Al2O3

Now to balance OO, the LHS has 2, while the RHS has 3, 2*3=623=6, so we have 2Al+3O_2>2Al_2O_32Al+3O2>2Al2O3

Mow AlAl is unbalanced again, to balance, we just add an extra 2 AlAl to the LHS to get 4Al + 3O_2>2Al_2O_34Al+3O2>2Al2O3.

Jul 23, 2017

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

Explanation:

2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)2Al(s)+32O2(g)Al2O3(s)

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

4Al(s) + 3O_2(g) rarr 2Al_2O_3(s)4Al(s)+3O2(g)2Al2O3(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_2H2,N2,O2,F2,Cl2 are BINUCLEAR.