What is the amount of limiting reagent in this chemical reaction? In the reaction involving $\pu{124 g}$ of aluminium and $\pu{601g}$ iron (III) oxide, what is the amount of substance in moles of the limiting reagent?


Here is what I have so far:

Balanced equation:
$$\ce{2Al + Fe2O3 -&gt; Al2O3 + 2Fe}$$

$\ce{Al}$ has MW = $27$, mass = $\pu{124 g}$ and determined its amount using $n=\frac{\text{mass}}{MW} = 4.59$

$\ce{Fe2O3}$ has MW = $160$, mass = $\pu{601 g}$ and determined its amount using n=mass/MW = 3.756

I was able to identify limiting reagent by:

g $\ce{Al}$ --> n $\ce{Al}$ -->n $\ce{Fe2O3}$--> g $\ce{Fe2O3}$

$124 g --s1-&gt; 4.59 --s2--&gt; 2.295 -s3--&gt; 367 g$

$S1 = n = 124/27 = 4.59$

$S2 = \frac{\text{want}}{\text{have}} \times \ce{Al} = 1/2 \times 4.59 = 2.295$

$S3 = m = n\times MW = 2.259 x 160 = 367 g$

Therefore as all $\ce{Al}$ did not use all the $\ce{Fe2O3}$ that means the $\ce{Al}$ is the limiting reagent.

The amount of substance in moles of limiting reagent $\ce{Al}$ (answer to the question) is $4.59$ OR $9.18$ (which is moles multiplied by the stoichiometric coefficient, which is $2$).
$4.59$ is not "something" (you haven't specified the unit) of an $\ce{Al}$ mole, but your amount of $124\mathrm{~g}$ of $\ce{Al}$ is exactly $4.59$ moles of $\ce{Al}$, it's just a unit change. In reaction you wrote, every two moles of $\ce{Al}$ reacts with one mole of $\ce{Fe2O3}$. This may help you to find the solution.