# Question #d5ff0

Jan 15, 2016

$4 F e + 3 {O}_{2} + 6 {H}_{2} O \rightarrow 4 F e {\left(O H\right)}_{3}$

#### Explanation:

We have metallic iron, reacting with oxygen gas and water
$F e + {O}_{2} + {H}_{2} O$

What would happen with a bar of iron, left outside (where the air is humid)? Rust, of course. Rust is the name of a bunch of stuff, the main ones being ferric oxide $F {e}_{2} {O}_{3}$ and ferric hydroxide $F e {\left(O H\right)}_{3}$, however as it turns out, the former is just what happens when you dehydrate the second, so we can say the reaction goes like

$F e + {O}_{2} + {H}_{2} O \rightarrow F e {\left(O H\right)}_{3}$

Which isn't quite balanced because of that pesky water, now thinking mathematically, we must have the same number of $H$s, $O$s and $F e$s on both sides.

If we have $x$ moles of water and $y$ moles of ${O}_{2}$ and $z$ moles of $F e$ we have

$2 x$ H on LHS, $x + 2 y$ O on LHS and $z$ Fe on LHS, to

$3 z$ H on RHS, $3 z$ O on RHS and $z$ Fe on RHS.

If we pick $z = 1$, we'll have

$2 x = 3 \rightarrow x = \frac{3}{2}$
$\frac{3}{2} + 2 y = 3 \rightarrow 2 y = 3 - \frac{3}{2} \rightarrow y = \frac{3}{4}$

$F e + \frac{3}{4} {O}_{2} + \frac{3}{2} {H}_{2} O \rightarrow F e {\left(O H\right)}_{3}$

Multiplying by $4$,

$4 F e + 3 {O}_{2} + 6 {H}_{2} O \rightarrow 4 F e {\left(O H\right)}_{3}$

Now, as the name implies oxygen oxidizes the other, because it likes to go from $0$ charge to $- 2$ charge, while it'll make the iron go from $0$ charge to $+ 3$ charge, which means Oxygen reduces itself and is the oxydizing agent, whereas Iron oxidizes itself and is the reducing agent.

This, like all redox equations can be accelerated by heat.