# Question #489d0

Oct 2, 2017

$2 : 1$

#### Explanation:

Yes, the mole ratio that exists between iron and iron(III) oxide can be written as $4 : 2$, i.e. $2$ moles of iron(III) oxide require $4$ moles of iron.

The idea here is that the mole ratio that exists between two chemical species that take part in a reaction is given by the coefficients added in the balanced chemical equation.

$\textcolor{b l u e}{4} {\text{Fe"_ ((s)) + 3"O"_ (2(g)) -> color(red)(2)"Fe"_ 2"O}}_{3 \left(s\right)}$

According to the balanced chemical equation, you have a $\textcolor{b l u e}{4}$ coefficient in front of elemental iron and a $\textcolor{red}{2}$ coefficient in front of iron(III) oxide.

This tells you that for every $\textcolor{b l u e}{4}$ moles of iron that take part in the reaction, the reaction produces $\textcolor{red}{2}$ moles of iron(III) oxide.

So you can say that iron and iron(III) oxide have a $\textcolor{b l u e}{4} : \textcolor{red}{2}$ mole ratio in this reaction.

You can divide both values by $2$ to simplify this mole ratio to

$\left(\textcolor{b l u e}{4} \textcolor{w h i t e}{.} {\text{moles Fe")/(color(red)(2)color(white)(.)"moles Fe"_ 2"O"_ 3) = "2 moles Fe"/("1 mole Fe"_ 2 "O}}_{3}\right)$

You can thus say that when elemental iron reacts with enough oxygen gas, the reaction consumes twice as many moles of iron as the number of moles of iron(III) oxide it produces.