Chapter 5: Unit 8. Limiting Reactant

Limiting Reactant

Often in a chemical reaction a reagent runs out and that stops the entire chemical reaction. Limiting reactant is a reactant that is completely used up first in a chemical reaction. The number of moles of limiting reactant determines the moles of products using the molar ratios in the balanced equation. The amount of products obtained from a limiting reagent is called theoretical yield. Theoretical yield is the max. amount of products theoretically possible in a chemical reaction according to balanced equation.

Go to the following simulation activity and find how may bread and cheese you need

https://phet.colorado.edu/sims/html/reactants-products-and-leftovers/latest/reactants-products-and-leftovers_en.html

The above picture shows that following the recipe we need 2 breads and 1 slice of cheese to make a product.

Now if you increase the number of cheese slices that will be excess reagent.

If you take less than one slice then there is no product because number of cheese slice is then limiting reagent and bread is excess.

Now we come to problems in chemical world. Can you explain the picture below?

Which one is limiting reagent? Which one is excess reagent? What is the theoretical yield of

the following reaction?

Example #1: Here’s a nice limiting reagent problem we will use for discussion. Consider the reaction:

2Al + 3I₂ ——> 2AlI₃

Determine the limiting reagent and the theoretical yield of the product if one starts with:

(a) 1.20 mol Al and 2.40 mol iodine.
(b) 1.20 g Al and 2.40 g iodine
(c) How many grams of Al are left over in part b?

Solution for part (a):

We already have moles as the unit, so we use those numbers directly.

1) Here is how to find out the limiting reagent:

take the moles of each substance and divide it by its coefficient in the balanced equation. The substance that has the smallest answer is the limiting reagent.

2) Let’s say that again:

to find the limiting reagent, take the moles of each substance and divide it by its coefficient in the balanced equation. The substance that has the smallest answer is the limiting reagent.

1) Resuming with the problem solution:

For aluminum: 1.20 / 2 = 0.60
For iodine: 2.40 / 3 = 0.80

2) The lowest number indicates the limiting reagent. Aluminum will run out first in part (a) of the question. Why?

1.20/2 means there are 0.60 “groupings” of 2 and 2.40/3 means there are 0.80 “groupings” of 3. If they ran out at the same time, we’d need one “grouping” of each. Since there is less of the “grouping of 2,” it will run out first.

3) The second part of the question “theoretical yield” depends on finding out the limiting reagent. Once we do that, it becomes a stoichiometric calculation.

Al and AlI₃ stand in a one-to-one molar relationship, so 1.20 mol of Al produces 1.20 mol of AlI₃. If molar mass of AlI₃ is 407.99 g/mol, the theoretical yield is 1.20 mol * 407.99 g/1 mol= 489.6 g = 490. g.

Notice that the amount of I₂ does not play a role, since it is in excess.

Solution for part (b):

1) Since we have grams, we must first convert to moles. The we solve just as we did in part a just above. For the mole calculation:

aluminum is 1.20 g / 26.98 g mol¯¹ = 0.04477 mol
iodine is 2.4 g / 253.8 g mol¯¹ = 0.009456 mol

2) To determine the limiting reagent:

aluminum is 0.04477 / 2 = 0.02238
iodine is 0.009456 / 3 = 0.003152

The lower number is iodine, so we have identified the limiting reagent.

3) Finally, we have to do a calculation and it will involve the iodine, NOT the aluminum.

I₂ and AlI₃ stand in a three-to-two molar relationship, so 0.009456 mol of I₂ produces 2/3*0.009456= 0.006304 mol of AlI₃. Again, notice that the amount of Al does not play a role, since it is in excess.

From here figure out the grams of AlI₃ and you have your answer.

Solution for part (c):

Since we have moles, we calculate directly and then convert to grams.

Al and I₂ stand in a two-to-three molar relationship, so 0.009456 mol of I₂ uses 0.006304 mol of Al.

Convert this aluminum amount to grams and subtract it from 1.20 g and that’s the answer.

Questions:

1. Find the limiting reagent in each of the following:
2. N₂ + 3H₂ → 2NH₃

Starting with 28.0 g of N₂ and 2.50 g of H₂

1. CaO + CO₂ → CaCO₃

Starting with 112.0 g of CaO and 66.0 g of CO₂

Ans: 1. a) H₂ b) CaO