Reactants, Products and Leftovers (Part II) [03 May, 2021]

STEM Chat Simulation Chat
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[size = 6]Reactants, Products and Leftovers[/size]

We discussed the concept of “Reactants, Products and Leftovers” from the scratch via PhET simulation in the Part I. We will extend our discussion here. :slight_smile:

Here our main focus is in the Level 2 and Level 3 of the Game section of the simulation.

Reflection

  1. Looking the balanced chemical equation how one will predict the amount of products and leftovers for a given amount of reactant?
  2. What is the limiting reagent? How the concept of limiting reagent is helpful in determining the amount of products?
  3. Given the balanced chemical equation, 4NH[size = 0.5]3[/size] + 3O[size=0.5]2[/size] ----> 2N[size = 0.5]2[/size] + 6H[size=0.5]2[/size]O
    Now if 6 molecule of NH[size = 0.5]3[/size] and 4 molecule of O[size=0.5]2[/size] is given as reactants, what will be the number of molecules of the products and leftovers? Which reactant will be limiting reagent in this case?
  4. How to determine the limiting reagent for a given reaction?
  5. Share your thoughts, questions and comments regarding the simulation and your experience during exploring the simulation.

We welcome all kind of suggestions and feedback if this was useful and how can be collectively make it better.

Acknowledgements
Discussion
Collaborators in video resource creation: Prof Savita Ladage, Deepak Arora, Hanza George, Mursaleen Shaikh, Ravi Sinha, Ashish Pardeshi, Indrani Das, Swarnava Mitra

Credits
PhET Interactive Simulations
University of Colorado Boulder
https://phet.colorado.edu

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@ravi312 asked Q3 in the discussion. We tried to find the limiting reagent in that case during the discussion session. But we won’t be able to come to any satisfactory answer.

I asked one of my friend @Hemant about the same. According to him, we will only call something limiting reagent when it completely vanished from the reaction medium. So for 2 component reaction if more that one component is left in the medium, the concept of limiting reagent won’t work there. Hence, it’s inconclusive in this case.

This seems reasonable to me. I will also urge others to look into it and put forward their comments on the same. @ravi312, @Ashish_Pardeshi @savital @mursaleen @indrani @deepak @hanza

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Hi !,
I would like to differ slightly, even if one of the reactant still does not vanish completely from the reaction, still the concept of limiting agent will hold. Example take this reaction-
1SO2 + 3H2 → 1H2S + 2H2O
and let us suppose, we have 6 molecules of SO2 and 4 molecules of H2O. So after the reaction, there will be 1of H2S, 2 of H2O formed as products and 5 of SO2 and 1of H2 remaining as leftovers.
Here, we see that the amount of H2 (which is less before the reaction) will determine the amount of products formed, although it is still remaining as a leftover. So, in this case, H2 is the limiting agent. (This is one of the example from Level 3 of phet simulation). We need at least 2 more of H2 for the next set of reaction to occur.

So, in bulk medium (not molecular level),which is the laboratory scale, you will not always be able to exactly say that one of the reactant is completely used up in a reaction, often that is determined by other factors such as the equilibrium constant value of that reaction, however, if the amount (or moles) of a reactant present in the system is less than the other (wrt to the stoichiometry of the reaction), then that reactant becomes the limiting reagent.

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Yes, I agree with you regarding this. But this is the limitation of simulation, that it is not talking about moles, it is talking about the molecules.

Regarding the example you gave, if you multiply the balance equation by 6, you can easily see that we need 18 H[size = 0.5]2[/size] molecules, for 6 SO[size = 0.5]2[/size] molecules. Even if you add only 2 molecules of H[size = 0.5]2[/size] you are able to get next set of products. So, here H[size = 0.5]2[/size] is the limiting reagent. Whereas, in the Q3, we neither multiply the equation by a natural number (not real as we are talking of molecules not moles) nor add only one component to the system to get next set of products, we need to add both components. So both the above reasons won’t valid in this case.

I guess this is the limitation of the simulation that it considers molecules not moles for easy representation purposes.

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Kindly find the note shared by Prof Savita Ladage (@savital ) on limiting reagent which will be useful for us to clarify the concept.

In sandwich example presented in the simulation- we started with the case where the product was defined and then at next level it allowed us to define product.

When sandwich was defined (e.g. two bread slice and 1 cheese slice together)

2\ bread\ slice + 1\ cheese\ slice = 1\ sandwich

If you have 4 bread slice and 2 cheese slice - the situation where both entities are multiplied by 2 - you will get only product and the concept of limiting is not valid for such case.
Now let us consider we have 4 bread slices and 3 cheese slices– how many sandwiches and any leftover? Which is limiting entity?
So the first entity is twice that of what is indicated in equation whereas the second entity is 3 thrice than that is indicated in the equation.

2\ bread\ slice + 1\ cheese\ slice = 1\ sandwich

First consider only bread slices and let us ask how $many sandwiches it can make? (Currently we are not paying attention to cheese slices.
Since we have 4 bread slices it is sufficient to make 2 sandwiches.
Now let us shift our attention to cheese slices
How many sandwiches will be obtained with 3 cheese slices – the answer is 3.

Now let us ask question - which entity is giving us minimum number of sandwiches? In other words- which is limiting the number of sandwiches?
The answer is entity is bread slices - So bread will be limiting reagent and it is totally consumed. So leftover is only cheese slices

Given the balanced chemical equation,

4NH_3 + 3O_2 ----> 2N_2 + 6H_2O

6NH_3 + 4O_2 ----> ?

Now if 6 molecules of NH_3 and 4 molecules of O_2 is given as reactants, what will be the number of molecules of the products and leftovers?
Which reactant will be limiting reagent in this case?
Again we have defined product and also proportion of each component of product
Now if 6 molecules of NH_3 - it can make how many entities on products in the proportion indicated?
We need to pick up 4 NH3 molecules each time. So by doing so we can make 2 molecules of N_2 + 6 molecules of H_2O. Now we have 2 NH_3 molecules which is not sufficient to make extra entities of product in the stated proportion (we are dealing with individual molecules here and thus fractional numbers cannot be considered)
Thus, if I have 6 NH_3 molecules- I can get 2 molecules of N_2 + 6 molecules of H_2O and 2 NH_3 molecules will be left over

Now let us look at O_2 molecules- by similar argument- once again we will get 2 molecules of N_2 + 6 molecules of H_2O and 1 O2 will be left over.

Which reactant is giving us less product? -In this situation, since both reactants are giving same number for products- for me- it is not possible to decide the limiting reagent. The limiting reagent needs to be consumed totally (please see the last para)

Now let us look at same equation but considering them as moles as in reality we deal with moles (mole is bunch of n entities and 1 mole of any substance has same fixed number of entities).
How the argument is then advanced?

4NH_3 + 3O_2 ----> 2N_2 + 6H_2O

4 moles of NH_3 will give 2 moles N_2 and 6 moles of H_2O
6 moles of NH_3 will give (6* 2)/4= 3 moles of N_2 and (6*6)/4=9 moles of H_2O

Now let us look at O_2

4 moles of O_2 will give (4*2)/3= 2.66 moles of N_2 (here fraction is allowed) and (6*4)/3=8 moles of H_2O

Which moles of reactants is giving you less moles of products?
The answer is Oxygen and thus oxygen will be limiting reagent. It will be consumed totally.

Those you are interested can read

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Limiting_Reagents

I quote from the above link

“The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. From the reaction stoichiometry, the exact amount of reactant needed to react with another element can be calculated. If the reactants are not mixed in the correct stoichiometric proportions (as indicated by the balanced chemical equation), then one of the reactants will be entirely consumed while another will be left over. The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant.”

(https://chem.libretexts.org/ - those who are interested please explore this link)

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