Question

What is the limiting reactant when 6.8 moles of iron react with 8.9 moles of oxygen gas?

Unbalanced equation: Fe + O2 \(\rightarrow\) Fe2O3

Answer 1

I am pretty sure that this is the balanced equation. \(\sf \color{red}{4}Fe+\color{red}{3}O_2\rightarrow \color{red}{2}Fe_2O_3\)

Answer 2

@iambatman

Answer 3

@Abhisar

Answer 4

Okay, so do you know what a limiting reagent is?

Answer 5

The reactant that is completely used first in a reaction

Answer 6

Exactly! You have to see, is it the oxygen or the iron?

Answer 7

Since they mentioned the moles, use moles to find out

Answer 8

Do you know how to continue?

Answer 9

Sorry, was afk. No, that is where I am getting stuck.

Answer 10

Alright. if you simply look at the numbers this problem is relative straight forward:
for the reaction to proceed we need 4 mol of iron to react with 3 mol of oxygen gas. If you simply "count the amount of times the reaction run" with each of the amount of substances we find the answer.

Lets start with Fe:
We need 4 mol the reaction to run ones:
6.8-4 = 2.6 mol left. aka Fe could run 1 time

oxygen:
3 mol used for one run.
8.9-3 = 5.9 for one, but it can run more time as we just need 3 as minimum.
5.9-3 = 2.9 for one more run. AKA you can run the reaction twice if oxygen was your limited reactant.

Out of those simple considerations you find the limiting reactant.

Answer 11

Fe would be the limiting reactant

Answer 12

As simple as that, yes

Answer 13

That makes it seem really easy :0

Answer 14

* :)

Answer 15

To make a more general approach:

Count the amount of times the reaction will run using:
\[\large \rho=\frac{ n(X) }{a }\]
where \(n(X)\) is the amount of substance of each molecule in the reaction and \(a\) is the stoichiometric number of the molecule in the reaction of interest, and \(\rho\) is the amount of "times the reaction ran".
if \(\rho\) equals the same for two or more molecules in the reaction, then you take the smallest \(n(X)\), as your answers, as moles is a huge number and it therefor make sense to talk about "half", "1/36" and ect. reactions.

Answer 16

And no problem at all. Thanks are to @aroub and @Abhisar

Answer 17

Wow @Frostbite
That's great explanation ...I couldn't have done better than this :)