How is the oxidation of aluminum metal represented?

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How is the oxidation of aluminum metal represented?

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Answer 1 · 2017-07-23T10:18:11

$4 A l + 3 O > 2 A l_{2} O_{3}$ $4Al+3O_">2Al_2O_3$

Explanation:

At first we have $A l + O_{2} > A l_{2} O_{3}$ $Al+O_2>Al_2O_3$

Let's first balance $A l$ $Al$ , the LHS has 1 Al atom, while the RHS has 2. To balance this we add another $A l$ $Al$ to the LHS, to get: $2 A l + O_{2} > A l_{2} O_{3}$ $2Al+O_2>Al_2O_3$

Now to balance $O$ $O$ , the LHS has 2, while the RHS has 3, $2 \cdot 3 = 6$ $2*3=6$ , so we have $2 A l + 3 O_{2} > 2 A l_{2} O_{3}$ $2Al+3O_2>2Al_2O_3$

Mow $A l$ $Al$ is unbalanced again, to balance, we just add an extra 2 $A l$ $Al$ to the LHS to get $4 A l + 3 O_{2} > 2 A l_{2} O_{3}$ $4Al + 3O_2>2Al_2O_3$ .

Answer 2 · 2017-07-23T10:21:02

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

Explanation:

$2 A l (s) + \frac{3}{2} O_{2} (g) \to A l_{2} O_{3} (s)$ $2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)$

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

$4 A l (s) + 3 O_{2} (g) \to 2 A l_{2} O_{3} (s)$ $4Al(s) + 3O_2(g) rarr 2Al_2O_3(s)$

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. $H_{2}, N_{2}, O_{2}, F_{2}, C l_{2}$ $H_2, N_2, O_2, F_2, Cl_2$ are BINUCLEAR.

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How is the oxidation of aluminum metal represented?

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