Copyright

Limiting Reactants & Calculating Excess Reactants

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: Calculating Reaction Yield and Percentage Yield from a Limiting Reactant

You're on a roll. Keep up the good work!

Take Quiz Watch Next Lesson
 Replay
Your next lesson will play in 10 seconds
  • 0:04 Limiting Reactant
  • 0:31 What Is the Limiting Reactant?
  • 1:24 Example One
  • 4:14 Example Two
  • 6:30 Lesson Summary
Save Save Save

Want to watch this again later?

Log in or sign up to add this lesson to a Custom Course.

Log in or Sign up

Timeline
Autoplay
Autoplay
Speed Speed

Recommended Lessons and Courses for You

Lesson Transcript
Instructor: Amy Meyers

Amy holds a Master of Science. She has taught science at the high school and college levels.

In this lesson, you'll learn about limiting and excess reactants and how to determine which reactant is the limiting one in a chemical reaction when given the amount of each reactant, and also how to calculate the amount of product produced.

Limiting Reactant

Pretend you're making chocolate chip cookies. You want to make a full batch of them (24). The recipe makes 24 cookies, but requires two cups of chocolate chips. You have enough of all the other ingredients but only have one cup of chocolate chips. So what ingredient is preventing you from making a full batch? The chocolate chips! You could make a batch with half the chips, but that wouldn't be any good. The lack of chocolate chips is limiting you in making a full batch. It is the limiting factor.

What Is the Limiting Reactant?

In chemistry, just like real life, there are rarely exact amounts of substances undergoing chemical reactions. There is usually too much of one reactant and not enough of another. The reactant that's used up first and so that no more product can be made is the limiting reactant. The substance that is in excess that doesn't get used up as a reactant is the excess reactant.

If you want to make CO2 and you start with C and O2, the reaction looks like this:

C + O2 --> CO2

One mole of carbon reacts with one mole of O2 to make one mole of CO2. What if I have two moles of C and one mole of CO2? Can I make two moles of CO2? No. Why not? Because I still only have one mole of O2, so I will run out of O2 before I run out of C. O2 is the limiting reactant. This lesson will teach you how to determine the limiting reactant in a reaction and calculate how much excess reactant you have.

Example One

Iron corrodes in the equation 3Fe + 4 H2 O --> Fe3 O4 + 4 H2. If I have 40g of water that react with 150g of iron, what is the limiting reactant? How much excess reactant do I have?

3Fe + 4 H2 O --> Fe3 O4 + 4 H2

You have 40g H2 O, and it reacts with 150g Fe. What is the limiting reactant and what is the mass in grams that is produced? Start with what you're given: 40g H2 O, 150g Fe, ???, Fe3 O4.

A previous lesson taught you how to determine how many grams of substance are in a mole of that substance. For this lesson, I will just tell you:

18g H2 O/1 mole, 55.8g Fe/1 mole, 232g Fe3 O4/1 mole.

What do you need to do first to determine the limiting reactant? You need to change the mass of the reactants to moles of reactants:

  • 40g H2 O x (1mole H2 O/18g H2 O) = 2.22 moles H2 O
  • 150g Fe x (1 mole Fe/55.8g Fe) = 2.69 moles Fe

Now let's pick one of the reactants to use in our comparison. It doesn't matter which one we pick. Let's use H2 O. We need to compare the moles of H2 O that we have to the mole ratio of Fe to H2 O. You can see that the moles of H2 O will cancel out and this will give us the moles of Fe that we need to use up all of the moles of H2 O that we have.

2.22 moles H2 O x 3 moles Fe/4 moles H2 O = 1.67 moles Fe.

So, for the 2.22 moles of H2 O that we have, we need 1.67 moles of Fe to use it all up in the reaction. Now we could make some nice rules for how to figure out the limiting reactant, but since you probably already have too many rules to learn, let's just think this through. We figured out earlier that we have 2.69 moles Fe available, and we now know that we only need 1.67 moles Fe to use up the 2.22 moles of H2 O that we have. So the H2 O will be used up before the Fe is gone. H2 O is the limiting reactant, the Fe is the excess reactant, and you will have 2.69 - 1.67 = 1.02 moles Fe left over.

To determine how much product Fe3 O4 will be made, multiply the limiting reactant times the mole ratio of product to the limiting reactant and then multiply by the molar mass of the product. Again, you can follow along as the labels cancel out.

To unlock this lesson you must be a Study.com Member.
Create your account

Register to view this lesson

Are you a student or a teacher?

Unlock Your Education

See for yourself why 30 million people use Study.com

Become a Study.com member and start learning now.
Become a Member  Back
What teachers are saying about Study.com
Try it risk-free for 30 days

Earning College Credit

Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.

To learn more, visit our Earning Credit Page

Transferring credit to the school of your choice

Not sure what college you want to attend yet? Study.com has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.

Create an account to start this course today
Try it risk-free for 30 days!
Create an account
Support