# Question d24f9

Apr 21, 2017

$\text{mol(O"_2")"~~1.51" moles}$

#### Explanation:

Rust is better assumed to be iron oxide, but nonetheless I shall answer this question.

First, consider the reaction that would be taking place. We're supposing that iron will react with water vapour and oxygen in the air to produce iron (III) hydroxide, which would probably look like this:

$4 {\text{Fe " + " "6"H"_2"O "+" "3"O"_2" "->" "4"Fe(OH)}}_{3}$

From this, we can start to approach this question by considering the mole ratio of components of this equation. To do that, we'll need to find the moles of iron we're starting with.

$\text{mol(Fe) " ="mass(Fe)"/("A"_r"(Fe)")" , where A"_r" is the relative atomic mass of iron, 55.8.}$
$\text{mol(Fe) } = \frac{112}{55.8} = 2.0072$

Now, we return to the equation we constructed earlier: How much oxygen do we need to react with all of the iron we have? Well, reacting $4$ moles of iron requires $3$ moles of oxygen, so we'll step the ratio through to find the moles of oxygen consumed.

"Fe"/("O"_2) = 4/3 = 1/(3/4) = 2.0072/1.5054#

So approximately $1.51$ moles of oxygen are required for this process to take place.