What is a Limiting Reagent?

In a chemical reaction, the Limiting Reagent, also known as the "limiting reactant", is the substance that is totally consumed when the chemical reaction is complete (1)

Example with the smaller compound as the limiting reagent: "
You are making cookies. The recipe you are using calls for 4 eggs for every 6 cups of sugar. You look in your kitchen and find 20 eggs and only 3 cups of sugar. You realize you can only make as many cookies as 3 cups of sugar allows, because if you used all the eggs first you wouldn't have nearly enough sugar to acurately fulfill the recipe. The cookies would be ruined! What you have to do is use up the 3 cups of sugar and then only 2 of your 20 eggs to follow the ratio of the recipe and make a good
 batch of cookies! Sugar will run out first, it is the limiting reactant! (important note: the compound present in the smallest amount is not necessarily the limiting reactant)." Site (

Example with the larger compound as the limiting reagent: In a bike store, a man is handmaking his own bikes for sale. Every bike needs 2 wheels and 1 horn. The man has 20 wheels and 14 horns in stock. The man can make 10 bikes using the 20 wheels he has in stock, but he can make 14 bikes with the horns in stock. Using both though, the man can only make 10 bikes because then there will be not enough wheels, therefore the wheels is the limiting reagent even though there are more tires then horns in stock.

This all might make sense when talking about recipes and bikes but how can this done usually chemicals and chemical reactions?
The way to find the limiting reagent of a compound is not hard, it just takes some practice.
Here is a problem

For the balanced equation shown below, what would be the limiting reagent if 95.0 grams of C2H4O were reacted with 306 grams of O2? (5)

2C2H4O+5O2 Yields 4CO2+4H20

Define yield
Step 1. Find the
GFW (molar mass) of the first and second compounds
For those of you who do not know how to locate the molar mass, I will give a quick tutorial
First find the atomic masses of all of the elements within each compound
Carbon has an Atomic mass of 12.011 which we can round to 12
Hydrogen has an Atomic mass of 1.00795 which we can round to 1
Oxygen has an Atomic mas of 15.9994 which we can round to 16
Ignoring the coieffients at the moment and Multiply the atomic mass by the subscript number given
There are 2 Carbon, 4 hydrogen, and 1 Oxygen molecules present in the first compound
                   2 (12)   +    4 (1)            +       1 (16) = 44
The GFW of the first compound = 44
Then calculate the GFW of the second compound given
Can you do it without the steps?
Try it by your self (Explaination located in Glossary)
Answer = 32
Step 2. Once you know the GFW's of both of the compounds located at the start of the equation, find the ratio of the compounds aka the coiffienct of the two compounds
Coiffient of Compound One = 2
Coiffient of Compound Two = 5
The ratio of the compounds is
Coiffient of 1 : Coiffient of 2
Step 3. Find the # of moles within both compounds
# of moles = grams/ GFW
Compound 1
grams = 95
GFW = 44
grams/GFW = 95/44
# of moles for compound 1 = 2.15909090
Compound 2
grams = 306
GFW = 32
grams/GFW = 306
# of moles = 9.5625
Step 4. Multiply the # of moles by the ratio
It is easier to have the ratio as one two something else such as 1:3 or 1:4 so that you only have to multiply one of the compounds
I like to change the ratio but it is not nessisary
2:5 = 1:2.5
2.5(molarity 1) and 1(molarity 2)
2.5 (2.15) and 1(9.5625)
5.3975 < 9.5625
This makes compound 1 the limiting reagent or C2H4O

I'll help you one more time
For the balanced equation shown below, what would be the limiting reagent if 57.2 grams of C2H4 were reacted with 246 grams of O2?

C2H4+3O2 Yields 2CO2+ 2H2O

1) Find the GFW of the first two compounds at the start of the equation 
So you must find the atomic number of the elements 
Carbon = 12
Hydrogen = 1
Oxygen = 16 
Ignoring the coiffients, find GFW by multiplying the atomic number by the number in the subscript 
First Compound
2(12) + 4(1) = 28
GFW = 28
Second Compound 
2(16) = 32 
GFW = 32 
2) Then find the ratio of the two compounds or the ratio between the two coiffients 
The first compound does not have a coiffient that means that it is 1
The Second compound has a coiffient of 3
Therefore the ratio of the two compounds is 
3) Then figure out how many moles are in each compound
# of moles = grams/GFW
First compound
# of moles = grams/GFW
57.2 grams/28 = moles
moles = 2.04
Second Compound 
# of moles = grams/GFW
246 grams/32 = moles
moles = 7.687
4) Then multiply the moles by the ratio 
3(2.04) and 1(7.687)
Therefore the first compound is the limiting reagent

The limiting reagent (6)

You get it?
Click the arrow to practice 
some practice problems