Some of the content of this guide was modeled after a guide originally created by Openstax and has been adapted for the GPRC Learning Commons in April 2021.
This work is licensed under a Creative Commons BY NC SA 4.0 International License.
The limiting reagent in a chemical reaction is the substance that gets completely consumed first. An introduction to limiting reagents is given in the Stoichiometry section. This section includes an example with the steps that will help you find the limiting reagent of a chemical reaction.
Example 1
In the following reaction, 2.25 g of NH_{3} is allowed to react with 2.50 g of O_{2. }With that information, answer the following questions:
NH_{3} + O_{2} NO + H_{2}O
Step 1
Find the mols of O_{2 }and NH_{3 }using the following formula:
Finding the limiting reagent:
Substance 
Coefficient from the balanced equation 
Number of mols available 

Limiting reagent 
O_{2} 
1 
0.078 mols 

The one with the smallest ratio becomes the limiting reagent. In this case, it’s O_{2 }with the smallest ratio of 0.078 
NH_{3} 
1 
0.132 mols 

Step 2
Finding the amount of NO produced using the number of mols of O_{2 }since it’s the limiting reagent and it’s going to dictate how much NO can be produced.

NO 
O_{2} 


“Have” comes from the coefficient of balanced chemical equation 
Solving for x in the above equation gives 0.078 mols for x. Therefore, the amount of NO produced is 0.078 mols  with that, we can calculate the amount of NO produced in grams using the following formula:
Molar mass of NO = 30.01g/mol
Step 3
Finding the amount of NH_{3} that’s left. In order to find the amount of NH_{3} left, we first have to find the amount of NH_{3} used. It can be found using the molar ratios, as shown below:

NH_{3} 
O_{2} 


“Have” comes from the coefficient of balanced chemical equation 
Solving for x in the above equation gives 0.078 mols for x. Therefore, the amount of NH_{3} used is 0.078 mols, and the amount of NH_{3 }left is found as follows:
NH_{3} left = 0.132 mols – 0.078 mols = 0.054 mols
The mass of NH_{3} left can be found using the following equation: