How is the oxidation of aluminum metal represented?

2 Answers
Jul 23, 2017

Answer:

#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#.

Jul 23, 2017

Answer:

#"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.