Question

Consider the reaction Mg(s) + I2(s) -> MgI2(s)
Identify the limiting reagent if there are 100 atoms of Mg and 100 molecules of I2

Answer 1

We need to find which reactant, Mg or I, will make the smallest number of moles of MgI2.

Answer 2

Let's start with Mg.

Answer 3

First, turn the 100 atoms of Mg into moles of Mg. Let me know if you need help to do that.

Answer 4

That's the 6.022*10^23 thing right?

Answer 5

Yes.
\[100 atoms Mg \times \frac{ 1 mol }{ 6.02^{23} atoms Mg} \]

Answer 6

Okay done with that, got 1.66*10^-22 mol

Answer 7

My bad for taking so long. I was checking

Answer 8

It's fine. Go on:)

Answer 9

Now, see how many moles of MgI2 you get. Let me know if you need help with that.

Answer 10

Do the same for I2

Answer 11

Yeah, I'm having trouble with that.

Answer 12

Alright, the equation states that there is a 1:1 relationship with Mg and MgI2

Answer 13

It will look like this,

\[1.66^{-22} moles of Mg \times \frac{ 1 mole MgI2 }{ 1 mole Mg} = \]

Answer 14

So it will be \[1.66^{-22} moles of MgI2\]

Answer 15

Do the same for I2. Let me know if you need help with that.

Answer 16

What's different in finding the I2 in mol since it's molecules not atoms?

Answer 17

Not much. Remember\[1 mole = 6.02^{23} atoms, molecules, formula units, etc\]

Answer 18

Oh okay. I'll find that then.

Answer 19

Wait shouldn't it just be the same thing then?

Answer 20

So try it like this

\[100 molecules of I _{2} \times \frac{ 1 mole of I2 }{ 6.02^{23} molecules of I2} = \]

Answer 21

Yes, but you have to do the boring stuff like the rest of us

Answer 22

Did that. Agreed, it is boring.

Answer 23

It's the same thing as the Mg

Answer 24

Yep. So what now?

Answer 25

Was there any more information?

Answer 26

Ummm... it just said to find the limiting regent for it. That's all.

Answer 27

Ok, I say we figure which reactant has the smallest mass then.

Answer 28

Okay then. That's the Mg then.

Answer 29

I would go with that.

Answer 30

I hope I didn't waste your time. Good luck.

Answer 31

No but what do I do with that information? What's the regent?
Sorry, I'm pretty new to this. It's the summer assignment.

Answer 32

That's cool. I say which ever has the smallest mass will run out first and thus, is the limiting reactant.

Answer 33

I could be wrong

Answer 34

It's all I have right now. I can check the answer but I still don't know how to get it. Thanks, still.

Answer 35

It just says "stoichiometric mixture"

Answer 36

@asib1214 what do you think?

Answer 37

I googled that, and it said that a stoichiometric mixture is a reaction with no excess reactants. In other words, there is no limiting reactant. Sorry about the drawn out process.

Answer 38

It's fine. I at least somewhat understand it now. It's a start. Thanks so much!

Answer 39

No prob. Thank you too

Answer 40

There actually isn't a limiting reagent in this case unless the question is worded incorrectly.

Answer 41

According to the reaction Mg (s) + I2 (s) -> MgI2 (s), one magnesium atom reacts with every one I2 molecule to produce 1 molecule of MgI2. Since Mg and I2 react in a 1:1 ratio, you simply need equal amounts to Mg and I2 to not have any shortage of reactants. In this question, you do have equal amounts: 100 atoms of Mg for 100 molecules of I2. There is no excess or shortage—all of the Mg and I2 get used up.

Answer 42

@Rogue is right

Simply look at the stoichiometric numbers : you can read them as moles or as microscopic units (atoms, molecules, ions...).
Since the numbers are 1 and 1, then 100 and 100 represent a stoichiometric mixture.