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In the reaction involving $\pu{124 g}$ of aluminium and $\pu{601g}$ iron (III) oxide, what is the number of 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 = 124 g and determined moles using n=mass/MW = 4.59

$\ce{Fe2O3}$ has MW = 160, mass = 601 g and determined moles 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 number of moles of limiting reagent Al (answer the question) is 4.59 OR 9.18(which is moles multiplied by the stoichiometric coefficient, which is 2).

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    $\begingroup$ 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?!? $\endgroup$ – MaxW Mar 17 '16 at 21:19
  • $\begingroup$ 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... ? $\endgroup$ – Lighthart Mar 17 '16 at 21:53
  • $\begingroup$ 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. $\endgroup$ – user27980 Mar 17 '16 at 22:01
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$4.59$ 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.

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Just calculate which reactant gives the lower amount of Al2O3.

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.

Check out this site for tutorials on limiting reagents: http://www.grandinetti.org/limiting-reagents

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