February 6th - Limiting Reagent

Today we learned about limiting reagents or limiting reactants. We learned, that usually one chemical gets used up before the other. The one used up first will “limit” the reaction and the reaction will stop. By knowing the LR, we can determine the quantity of products formed. We already posses all the tools, it is just a matter of applying them!

Ex. A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. A) Which is the limiting reactant B) How much (in grams) nitrogen monoxide is formed and C) how much excess reactant remains after the reaction has stopped?

A) As we learned in class, the first step is to draw a free body diagram, as shown below:

4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)

Now we pick a reactant. I like ammonia, so we’ll use it. Figure out how much of ammonia we have and how much is required. By comparing these two values, it will be evident which reactant limits the reaction.

2.00g NH₃ x (1 mol NH₃ / 17.0 g) = 0.118 mol NH_{3 (present)}

4.00g O₂ x (1 mol O₂ / 32.0 g) x (4 mol NH₃ / 5 mol O₂) = 0.100 mol NH_{3 (needed)}

Since we have more ammonia than we need, oxygen is our limiting reactant.

B) Having determined the LR, we can now proceed with further calculations. To find the number of NO produced:

4.00g O₂ (1 mol O₂ / 32.0g) (4mol NO / 5 mol O₂) (30.0 g/ 1 mol NO) = 3.00 grams NO

C) We simply take the amount of ammonia present minus the amount needed.

0.118 mol NH₃ – 0.100 mol NH₃ = 0.018 mol NH₃ (17.0 g / 1 mol) = 0.306 g NH₃

And that’s it! So remember, when two masses of reactants are provided in the question, one of them is the limiting reactant. You’ll have to find it, unless one reactant is in excess.

To learn some more, watch the video below:

Posted by Andrew.