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: $$\ce{2Al + Fe2O3 -> Al2O3 + 2Fe}$$
$\ce{Al}$ 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 $\ce{Fe2O3}$
$124 g --s1-> 4.59 --s2--> 2.295 -s3--> 367 g$
$S1 = n = 124/27 = 4.59$
$S2 = \frac{\text{want}}{\text{have}} \times \ce{Al} = 1/2 \times 4.59 = 2.295$
$S3 = m = n\times MW = 2.259 x 160 = 367 g$
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 $2$).