# 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: $$\ce{2Al + Fe2O3 -> Al2O3 + 2Fe}$$

$$\ce{Al}$$ 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 $$2$$).

• If you only start with 4.59 moles of Al, how are you going to react 9.18 moles of Al with the iron (III) oxide?!?
– MaxW
Mar 17, 2016 at 21:19
• I think you have written the wrong number for moles of rust. Regardless of the typo, the number of moles you have is the number of moles you have. It is not doubled by coefficients. The coefficient means you need twice as many moles of aluminum as rust for a complete reaction. So... ? Mar 17, 2016 at 21:53
• I am confused as 1 mole of Al = 4.59 but we start with 2 molecules of Al, so don't we start with 9.18 mole of Al? This is my confusion about this. *Thanks for the typo I changed it. Mar 17, 2016 at 22:01

$4.59$ 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.
From Al you have $n_\text{Al$_2$O$_3$} = \displaystyle \frac{124 \text{ g Al}}{26.9815386 \text{ g/mol}} \cdot \frac{1\text{ mol Al$_2$O$_3$}}{2 \text{ mol Al} }= 2.30 \text{ mol}$,
From Fe$_2$O$_3$ you have $n_\text{Al$_2$O$_3$} = \displaystyle \frac{601 \text{ g Fe$_2$O$_3$}}{159.6882 \text{ g/mol}} \cdot \frac{1\text{mol Al$_2$O$_3$}}{1 \text{ mol Fe$_2$O$_3$} }= 3.76 \text{ mol}$.
Clearly Al is the limiting reagent with $n_\text{Al} = \displaystyle \frac{124 \text{ g Al}}{26.9815386 \text{ g/mol}} = 4.60$ mol.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}