# How do we determine the limiting reagent, and the reagent in excess?

Jan 16, 2014

You find the limiting reagent by calculating and comparing the amount of product each reactant will produce.

Let’s look at respiration, one of the most common chemical reactions on earth.

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

What mass of carbon dioxide forms in the reaction of 25.0 g of glucose with 40.0 g of oxygen?

Solution

Step 1: Write the balanced chemical equation for the chemical reaction.

Step 2: Convert all masses into moles.

25.0 g C₆H₁₂O₆ × (1 mol C₆H₁₂O₆)/(180.2 g C₆H₁₂O₆) = 0.139 mol C₆H₁₂O₆

40.0 g O₂ × (1 mol O₂)/(32.00 g O₂) = 1.25 mol O₂

Step 3: Calculate the mass of product from each reactant.

0.139 mol C₆H₁₂O₆ × (6 mol CO₂)/(1 mol C₆H₁₂O₆) × (44.01 g CO₂)/(1 mol CO₂) = 36.6 g CO₂

1.25 mol O₂ × (6 mol CO₂)/(6 mol O₂) × (44.01 g CO₂)/(1 mol CO₂) = 55.0 g CO₂

Since glucose gives the smaller amount of product, it is the limiting reactant.

Here is a video which uses this technique to solve a different limiting reactant question.

Please watch the video I've posted on YouTube. It describes a method you can use to determine limiting and excess reagents.

One way to do this is to calculate the amount of product that would be produced based only off of the starting amount of one reactant and then to calculate the amount of product that would be produced based off of the other reactant.

The calculation that produces the smaller amount of product reveals the reactant that will limit the amount of product that will be formed.

This strategy takes a bit longer than some techniques, but is a good way to really understand this concept.

Here is the video example of this technique:

Hope this helps!
Noel P.