[nozo]

Hi, I'm new to the forums.. and to Chemistry (let's just say I went back to school for a change in career) I'm sure I'll visit often as I'm planning on taking a lot of Chem courses  

Anyway, I'm a little lost on the concept of moles and limiting reagents... I know how to do the calculations, but some of the problems I'm not sure why this was the limiting reagent, etc.

Example, XX + YY --> ZZ
Now we add .007 L of XX with a molarity of 4.11
Mix with .005 L of water, and .009 L of YY with a molarity of 5.14
Density for all is 1g/ml

First thing I do is solve for the mols for XX, YY and water
mol XX = .007L(4.11) = .0288 mol XX
mol YY = .005L(5.14) = .0463 mol YY
mol water = 5g/18g = .278 mol water

So anyway I guessed that the limiting reagent was XX, since I just took the smallest mole value (the answer is right). But, I want to know why is it so? Searched here, and found something on molar ratio.. but in this equation, it doesn't say.. or is it assumed that the molar ratio is 1:1? I appreciate it if someone would explain it to me (in layman's terms even better )

Tia!

[AWK]

OK, but do not take sovent (water) into account for calculation of moles. Volume of water is eventually needed for calculation of concentrations.

[Borek]

Imagine you are making phones. Phone is made of 1 frame, 1 handset and 10 buttons. You have 5 frames, 4 handsets and 33 buttons. How many phones can you make?

Reaction equation is

frame + handset + 10 buttons -> phone

(note that chemically speaking we should write phone as FrameHandsetButton₁₀

Reaction equation tells you that 1 frame reacts with 1 handset and 10 buttons, or that 1 mole of frames reacts with 1 mole of handests and 10 moles of buttons.

You can make three phones, as you don't have enough button for more. Buttons are limiting reagent, and the ratios substances react is defined by the reaction equation.

[Hunt]

It's quite simple if you know that moles represent atoms/molecules. In your example, XX molecules are present in less amount than YY molecules. Therefore, it's logical to assume that XX will run out first & on this basis it is considered a limiting reagent.

To understand better, you might wanna consider a rxn with different stoichiometry, like
EDIT : 3 X + 2 Y ----> X₃Y₂ , and try to figure out  the limiting reagent in this case.

[Borek]

like
3 X + 2 Y ----> X₃ + Y₂ , and try to figure out  the limiting reagent in this case.

Looks like two separate reactions to me

[Hunt]

Thanks for the remark Borek. It seems I added by mistake the '+'.

[nozo]

Thanks everyone! This really helps

OK.. lemme see if I get it...

To understand better, you might wanna consider a rxn with different stoichiometry, like
EDIT : 3 X + 2 Y ----> X₃Y₂ , and try to figure out  the limiting reagent in this case.

.02877 mol X (2 mol Y / 3 mol X) = .0192 mol Y
.0463 mol Y (3 mol X / 2 mol Y) = .0695 mol X

So in this case..  Y is now the limiting reagent cuz like you said, it runs out quicker, right?

[Hunt]

I cant seem to understand how you came up with that answer. You've got to understand the concept, not just a formula/method of working. Let's start from scratch :

n = N / N_A

n : number of moles
N : number of Molecules
N_A : Avogadro's number

Hence, n_X / n_Y = N_X / N_Y ( N_A cancels )

Let's 1st consider that Y is the limiting reactant. In this case, all of Y (0.0463 mol) must react completely with X. Hence n_Y-rem = 0 mol . How much of X must react and how much remains?

n_X-react = 1.5 molecules/molecules x 0.0463 mol = 0.0693 mol

How much of X remains?
n_X-rem = n_X-initial - n_X-react = 0.0288 mol - 0.0693 mol = - 0.0405 mol

This is impossible! The amount of X remaining is a -ve value.Therefore, Y cannot be the limiting reactant.



Now, Let's consider that X is the limiting reactant. In this case, all of X (0.0288 mol) must react completely with Y. Hence n_X-rem = 0 mol . How much of Y must react/remain?

n_Y-react = 0.66 molecules/molecules x 0.0288 mol = 0.0192 mol of Y .

Therefore, 0.0192 mol of Y reacts with X , but how much of Y remains?

n_Y-rem = n_Y-initial - n_Y-react = 0.0463 - 0.0192 = 0.0271 mol

This is acceptable. Therefore, X is the limiting reactant



Instead of doing all this procedure, however, you can always know quickly which reactant is the limiting reagent.

Divide n_X by its stoichiometry : n_X / 3 = 0.0288 / 3 =  9.6 x 10^-3mol X

Divide n_Y by its stoichiometry : n_X / 2 = 0.0463 mol / 2 =  23 x 10^-3mol Y

Obviously, X is less and hence is the limiting reactant. This method is quicker especially when you're dealing with many reactants in a chemical reaction.

The important thing is not to get the right answer but rather to understand why.

[nozo]

Ah ok, thanks.. I think I understand it now!
Anyway, I'll go and do more problems...  
Thanks again!