When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 12.9 mol of iron in the rusting reaction? 4Fe(s) + 3O_2(g) -> 2Fe_2O_3(s)?

1 Answer
Mar 28, 2018

$9.675 \setminus \text{mol}$ of oxygen gas.

Explanation:

We have the balanced equation:

$4 F e \left(s\right) + 3 {O}_{2} \left(g\right) \to 2 F {e}_{2} {O}_{3} \left(s\right)$

And so, $4$ moles of iron react with $3$ moles of oxygen, and therefore, $12.9 \setminus \text{mol}$ of iron would react with:

12.9color(red)cancelcolor(black)("mol" \ Fe)*(3 \ "mol" \ O_2)/(4color(red)cancelcolor(black)("mol" \ Fe))=9.675 \ "mol" \ O_2