How to determine the limiting reactant???? [ENDORSED]

[Avery Zuelch 1D]

Hi guys! So in reviewing for the first test, I have found myself stuck on trying to determine the limiting reactant. I have tried numerous times attempting what Dr. Lavelle did in class to determine it, but it just isn't clicking. Does anyone have any advice??

[Julia Go 2L]

Hi Avery!

A good place to start is to make sure that your chemical equation is balanced. Next, figure out how many moles you have for each reactant by dividing the known masses of reactants by the molar mass of each reactant. After you figure out the numbers of moles you have of each reactant, look back to the balanced equation.

Compare the stoichiometric ratio of the reactants to the actual molar ratio of the reactants. You should be able to tell which reactant will "run out" first by looking at the stoichiometric ratio from the balanced equation.

Another method you can use is a guess and check type of method by testing all of the reactants as the limiting reactant. The reactant that yields the smallest amount of product will be the limiting reactant.

[Sophia Ding 1B]

It also took me some time to really understand it; personally I like to just take each reactant and calculate the moles or grams of the product (depending on what the question wants for the final answer) to see which one produces the least amount of product as that would be your limiting reactant. But I know that that requires much more work than need be, so once you hammer down that concept it's much easier to look at the molar ratio of the reactant to the product in terms of the balanced equation, as then you can kind of use your intuition to see which reactant would be the limiting one in the equation!

[Peichung Chou 1A]

The way I think about it is for each molecule you're testing in candidacy for limiting reagent, pretend the other is unlimited in amount. Calculate the molar ratio for the limited molecule with the product and see how much product you get. Do the same with the other molecule, then compare which molecule produced the least of a product.

[Nicole Lee 4E]

I usually just test all of the reactants to see how much I get of one specific product. Whichever reactant gives me the least product (mol or grams) is the limiting reactant. It is sometimes inefficient when you're only asked to find the limiting reactant, but most problems I've run into end up asking for the amount of product produced anyway.

[anishathomas]

The way I calculate the limiting reactant is by first finding the amount of moles are in the reactants given. Once you find the moles, only convert one of them to the moles of the other reactant. For example, if you had a equation of 2h2+2o2----2h202 find the moles of h2 through o2 by multiplying the moles of o2 found to the moles of h2 over the moles of the coefficient of o2 in the balanced equation. Once you find the moles of h2, you can compare that to the moles of h2 you have, with the amount of h2 needed, and if there is more than what's needed then that's you're excess reagent, and the o2 would be your limiting(vice versa as well).

[Rylee Nelsen 3A]

If you're having trouble, I would recommend watching this video https://www.youtube.com/watch?v=Mlu_v8rE1TY. It explains the process very clearly, and lets you see the organization of a limiting reactant problem! Hope that helps!

[Shibhon_Shepard]

How I understood this was once you fin the amount in grams, the one with the bigger amount of moles that is need for the reaction is the limiting reactant. this is because you need more of this reactant to complete the equation.

M9 in the textbook wasn't too much if you wanted a better idea with smaller numbers.

[Sameen Mahmood 3D]

Another method would be to find the moles of each reactant and compare the ratio of the calculated moles to the reaction's moles. For ex, if you have 1 mol C and 2 mol H, but the reactants' side is something like C+H, you know the 1:1 ratio isn't met. H is present in excess (you have more than what you would need as dictated by the ratio) so C is the limiting reactant. I hope that helped!!