Limiting Reagent

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 NH₃ is allowed to react with 2.50 g of O_2. With that information, answer the following questions:

NH₃ + O₂  NO + H₂O

1. In the above reaction, what is the limiting reagent? 
2. How many grams of NO are formed?  
3. How much of the excess reactant remains after the completion of the reaction?  

Step 1

Find the mols of O₂ and NH₃ using the following formula:

Finding the limiting reagent:

Step 2

Finding the amount of NO produced using the number of mols of O₂ since it’s the limiting reagent and it’s going to dictate how much NO can be produced.

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 NH₃ that’s left. In order to find the amount of NH₃ left, we first have to find the amount of NH₃ used. It can be found using the molar ratios, as shown below:

Solving for x in the above equation gives 0.078 mols for x. Therefore, the amount of NH₃ used is 0.078 mols, and the amount of NH₃ left is found as follows:

NH₃ left = 0.132 mols – 0.078 mols = 0.054 mols

The mass of NH₃ left can be found using the following equation: