## What is a Limiting Reagent?

The limiting reagent is simply the chemical that determines how far the chemical reaction will go before the amount of this chemical gets used up, causing it to stop or limit the reaction.

## How does this pertain to Chemistry and Everyday Life?

In every chemical equation there must be a proportion, the chemical which has less moles than is required by this proportion is known as the limiting reagent. Limiting reagents occur in all chemical reactions, making it an important element of Chemistry. In ones everyday life limiting reagents can be found when for example you have 4 hot dogs and 3 hot dog buns...the limiting reagent here would be the amount of buns because its limiting this "reaction."

## How to Identify the Limiting Reagent...

There are a couple of different ways to find the limiting reagent in a chemical equation. This is the method that I use in order to do so.

Here is an example:

For the balanced equation shown below, what would be the limiting reagent if 72.4 grams of C2H3OCl were reacted with 22.4 grams of O2?

C2H3OCl+O2 => 2CO+H2O+HCL

Step 1: Find the Molar Mass of each element

This can be done by going to the periodic table of elements and rounding the atomic mass to the nearest whole value (except for Cl, which is 35.5) http://www.webelements.com/

Since there are 2 Carbons you multiply the molar mass of C by 2...since there are 3 Hydrogens you multiply the molar mass of H by 3...and so on. Add them up at the end.

C2H3OCl: O2:

12 * 2 = 24 16 * 2 = 32

1 * 3 = 3 ______+_

16 * 1 = 16 32

35.5 * 1 = 35.5

_________+_

78.5

Step 2:Find the moles

This can be done by dividing the amount of grams given in the original question by the molar masses.

Since there were 72.4 grams of C2H3OCl, you would divide its molar mass 78.5 by 72.4 grams. (72.4/78.5) = .92

Since there were 22.4 grams of O2, you would divide its molar mass 32 by 22.4 grams. (22.4/32) = .7

Step 3: Set up a ratio based upon two proportions

The 1st proportion you set up is based upon the coefficients of the left side of the equation. _C2H3OCl+_O2 => 2CO+H2O+HCL

Since in this equation there are no #'s before the equations, it means 1

Meaning your first proportion would be set up as such:

1

1

The 2nd proportion you set up is based on the # of moles. Set up this proportion according to the 1st proportion by matching the molar value with the ratio value. This isnt really seen here because the 1st proportion is simply 1

1

Your ratio should now look like:

1 .92

1 = .7

Step 4: Cross Multiply

When you cross multiply

1 .92

1 = .7

you end up with .7 * 1 = .7 and .92 * 1 = .92

Step 5: Identify the Limiting Reagent

The limiting reagent is the smaller value which in this case is .7 which we got from the moles of O2 making O2 the limiting reagent.

After going through this explanation a few times, refer to the practice problems page located here to put yourself to the test! Finding the Limiting Reagent Practice Problems

created by Faiz Khan