# Iron rusts according to the following reaction: 4Fe (s) + 3O_2 (g) + 4H_2O -> 2Fe_2O_3(H_2O)_2. If 1.4 liters of oxygen were added to excess water and iron at a pressure of 0.97 atms and a temperature of 55 C, how many grams of iron would rust?

Jun 13, 2018

#### Explanation:

Let us readdress it....iron metal is oxidized to $F {e}_{2} {O}_{3}$...

$2 F e \left(s\right) + 3 {H}_{2} O \left(l\right) \rightarrow F {e}_{2} {O}_{3} + 6 {H}^{+} + 6 {e}^{-}$ $\left(i\right)$

Dioxygen is reduced to water...

$\frac{1}{2} {O}_{2} + 2 {H}^{+} + 2 {e}^{-} \rightarrow {H}_{2} O \left(l\right)$ $\left(i i\right)$

And $1 \times \left(i\right) + 3 \times \left(i i\right)$ gives....

$2 F e \left(s\right) + \cancel{3 {H}_{2} O \left(l\right)} + \frac{3}{2} {O}_{2} + \cancel{6 {H}^{+}} + \cancel{6 {e}^{-}} \rightarrow F {e}_{2} {O}_{3} + \cancel{6 {H}^{+}} + \cancel{3 {H}_{2} O \left(l\right)} + \cancel{6 {e}^{-}}$

…. and we cancel to give...

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

And note that this equation assumes COMPLETE oxidation....iron oxide chemistry, i.e. rust chemistry is a very broad church...and many iron oxides and hydrous oxides are known. And so we calculate the moles of dioxygen gas....and plug this value back into the given equation....

${n}_{\text{dioxygen}} = \frac{P V}{R T} = \frac{0.97 \cdot a t m \times 1.4 \cdot L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \cdot 328.15 \cdot K} \cong 0.05 \cdot m o l$....and given the equation...$\frac{4}{3} \cdot \text{equiv}$ of iron metal will oxidize...approx. $4 \cdot g$..