# 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!

Replay
Your next lesson will play in 10 seconds
• 0:05 Limiting Reactant
• 1:36 Calculating Limiting Reactant
• 2:08 Example One
• 4:43 Example Two
• 6:36 Lesson Summary

Want to watch this again later?

Timeline
Autoplay
Autoplay
Create an account to start this course today
Try it free for 5 days!

#### 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 what limiting reactant and excess reactant mean and how to determine which reactant is limiting in a chemical reaction when given the amount of each reactant. You'll also discover how to calculate the amount of product produced.

## Limiting Reactant

Pretend you are 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.

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 is used up first and prevents more product from being 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, you start with C and O2. C + O2 --> CO2. One mole of carbon reacts with one mole of oxygen to make one mole of CO2. What if I have two moles of C and one mole of O2? 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 a limiting factor. This lesson will teach you how to determine the limiting reactant in a reaction and calculate how much excess reactant you have.

## Calculating Limiting Reactant

To determine the limiting reactant, take the following steps.

1. Choose one reactant and call it A.
2. Calculate the amount of moles of the other reactant (call it B) that are needed to use up all of reactant A.
3. Compare the calculated amount of B to the actual amount available. If more is required than is available, then reactant B is the limiting reactant. If less is required than is available, then reactant A is the limiting reactant.

Let's look at an example, and this will become clearer.

## Example One

Iron corrodes in the equation 3Fe + 4 H2 O --> F3 O4 + 4 H2 . If I have 40g of water that react with 150g iron, what is the limiting reactant? How much excess reactant do I have?

If 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 are given: 40g H2 O, 150g Fe, unknown quantity F3 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 F3 O4/1 mole.

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

• 40g H2 O x (1mole H2 O/18g H2 O) = 2.2 moles H2 O
• 150g Fe x (1 mole Fe/55.8g Fe) = 2.7 moles Fe

Next, compare the moles of H2 O to the mole ratio. 2.2 moles H2 O x 3 moles Fe/4 moles H2 O = 1.65 moles Fe. So, for every 2.2 moles of H2 O, you need 1.65 moles of Fe to fully react. Since you have 2.7 moles Fe available, H2 O is the limiting reactant.

To determine how much product will be made, multiply the limiting reactant times the mole ratio of product to limiting reactant by the molar mass of product. 2.2 moles H2 O x (1 mole F3 O4/4 moles H2 O) x (232g F3 O4/1 mole) = 127.6g F3 O4.

## Example Two

Zinc and sulfur react to form zinc sulfide. If 100g Zn react with 30g S8, what is the limiting reactant? And, how much product is made? You are given the following information: 8Zn + S8 --> 8ZnS, 65.4g Zn/1 mole Zn, 256.5g S8/1 mole S8, 97.5g ZnS/1 mole ZnS.

Change mass of A to moles of A and mass of B to moles of B:

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

### Register for a free trial

Are you a student or a teacher?
Back

Back

### Earning College Credit

Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 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.