Question

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.

Answer 1

Answer:

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

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)$

Explanation:

Solid iron + Oxygen gas  →  Iron(III) oxide

$Fe(s)+O_2(g)\rightarrow Fe_2O_3(s)$

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

$2Fe(s)+O_2(g)\rightarrow Fe_2O_3(s)$

Step 2 : Writer 3 in front of $O_2$ and 2 in front of $Fe_2O_3$ to balcne the oxygen atom.

$2Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)$

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

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)$

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

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)$

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