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When iron rusts, solid iron reacts with gaseous oxygen to form solid iron(III) oxide.Enter a balanced chemical equation for this reaction. Express your answer as a chemical equation. Identify all of the phases in your answer.

Respuesta :

Answer:

The balanced chemical reaction of rusting of solid is given as;

[tex]4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)[/tex]

Explanation:

Solid iron + Oxygen gas  →  Iron(III) oxide

[tex]Fe(s)+O_2(g)\rightarrow Fe_2O_3(s)[/tex]

Step 1 : Writer 2 in front of Fe(s)

[tex]2Fe(s)+O_2(g)\rightarrow Fe_2O_3(s)[/tex]

Step 2 : Writer 3 in front of [tex]O_2[/tex] and 2 in front of [tex]Fe_2O_3[/tex] to balcne the oxygen atom.

[tex]2Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)[/tex]

Step 3 : Now, iron on product side is more so, write 4 on the place 2 in front of Fe(s).

[tex]4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)[/tex]

The balanced chemical reaction of rusting of solid is given as;

[tex]4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)[/tex]

According to reaction , 4 moles of solid iron reacts with 3 moles of oxygen gas to give solid compound as product named iron(III) oxide or ferric oxide

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