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LIMITING REACTANT
Limiting Reactant – the reactant that is used up first in a chemical reaction. It limits and determines the maximum
amount of product that could be produced. Once the limiting reactant is used up, the reaction stops and the product can no longer be formed. The other reactants are said to exist in excess.
Let’s use an analogy to show you why this type of problem is important…
What if you had 30 hamburger patties and 100 hamburger buns. Could you make 100 hamburgers from these materials?
I think you get the idea! You could think of the patties as being the “Limiting Reagent.”
The same ides is present in chemical reactions Here is an example…
Suppose we mix 100g of Mg with 100g of HCl. How much MgCl 2 could be produced?
Mg + 2HCl → MgCl 2 + H 2 100g 100g
- Convert 100g of each chemical into moles. 100g Mg = 4.11 Moles of Mg 100g HCl= 2.73 moles of HCl
- From the balanced chemical equation look at the mole ratio. 1:2 Magnesium to Hydrochloric acid
- Can all 100g of Mg react completely? NO! From the balanced equation you can see that each mole of Mg needs 2 moles of HCl. This means that if we had 4.11 moles of Mg, then we would need 8.22 moles of HCl to react. We don’t have that much HCl.
- Looking at this from the other side, if we have 2.73 moles of HCl, then we can only react with half that much, or 1.36 moles of Mg. The rest of the Mg will be left unreacted.
Since the HCl limits the amount of Mg that can react, we call the HCl the Limiting reagent.
- Now that we have determined the actual amounts of chemicals that will react (2.73 moles HCl and 1. moles of Mg) we can do the problem exactly as we did the other stoichiometric problems. 1. The equation is already balanced. 2. The ratio of Mg to MgCl 2 is 1:1 so if we have 1.36 moles of Mg that can react, then we can make 1.36 moles of MgCl 2. 3. We can convert 1.36 moles of MgCl 2 into grams by multiplying the molecular weight of magnesium chloride, 95.3, we get 129.6 grams of MgCl 2 produced.
Can you see what will be left at the end of the reaction? All of the HCl will be gone. Only 1.36 moles of Mg can react, so there will be 4.11 – 1.36 = 2.75 moles of Mg left over.
Problem
Examine the reaction below. How many grams of NaF could be produced if you mixed 500 grams of Na with 300 grams of F 2? 2Na + F 2 → 2NaF
In any problem where you are given quantities of two chemicals that will react, one will be used up before the other. The chemical will limit the extent to which the reaction will go.
Think about this…
Suppose you want to cook out and have hot doges.
What if you order 10 lbs of hot dogs and 10 lbs of Buns. Do you think that you will have the right number of buns for hot dogs?
No! Because the buns and hot dogs each have different weights.
Do you see the analogy to chemical reactions? Do you think that it is likely that 10g of calcium should react perfectly with 10 g of sulfur?
Ca + S = CaS
No! For the same reason that the hot dog buns and hot dogs won’t match up… They have different weights!
One hot dog doesn’t weigh the same as one Bun… and one calcium atom doesn’t weigh the same as one S atom.
This is why we use Moles
Practice Problem
Examine the reaction below.
H 2 + S → H 2 S
How many grams of H 2 S can be produced by reacting 50 grams of H 2 with 50 grams of S?
- Some of the product may be lost while you are recovering it. (spills, etc.)
It is sometimes useful to determine what percent of the theoretical yield was actually obtained.
Percent Yield – The ratio of the actual yield to the theoretical yield converted into a percentage. It is a measure of
the closeness of the actual yield to the theoretical yield.
Percent Yield = Actual x 100 Theoretical
Example: 4.9 grams of magnesium is burned in oxygen. The actual yield of magnesium oxide is 6.5 grams. The Stoichiometry calculations give a theoretical yield of 8.1 grams of magnesium oxide. The percent yield is calculated as follows:
Percent Yield = 6.5 grams MgO x 100 = 80% 8.1 grams of MgO
This tells us that 80% of the amount of product expected was produced/ recovered. This also tells us that 20% of the product was lost.
It is also useful to know how to determine how much you messed up!! In other words, how far off was the actual yield from the theoretical yield?
Experimental Error – is a measure of how far off the actual yield is from the theoretical yield. (How much you
messed up!) Experimental Error = Theoretical - Actual x 100 Theoretical
Example: 4.9 grams of magnesium is burned in oxygen. The actual yield of magnesium oxide is 6.5 grams. The Stoichiometry calculations give a theoretical yield of 8.1 grams of magnesium oxide. The experimental error is calculated as follows:
Experimental Error = 8.1 g MgO - 6.5 g MgO x 100 = 20% 8.1 grams MgO
This tells us we messed up by 20% or we lost 20% of our product! It also tells us that we produced or recovered 80% of what we should have.
HEAT IN REACTIONS
Exothermic – Heat is given off by a reaction and can be expressed on the product side.
Example: Burning methane in a Bunsen burner.
Endothermic – Heat must be supplied for a reaction to occur and is shown on the reactant side.
Example: The decomposition of potassium chlorate.
