There are different methods to find your limiting and excess reagents. We will show you 2 different methods. An excess reagent is the substance that you have to much of, and your limiting reagent is what you have to little of or what is limiting you. You will need stoichiometry to do these problems so make sure you understand that first.
Basic Example
cookie recipe:
2 eggs+ 2 cups of flour ------> 1 batch
At home I have 6 eggs and 10 cups of flour. How many batches of cookies can I make?
6 eggs x 1 batch/ 2eggs = 3 batches (maximum you can make)
10 cups of flour x 1 batch/2 cups of flour = 5batches
The eggs are the limiting reagent because whether you have 10 cups of flour of 100 cups of flour you only have enough eggs to make 3 batches. The cups of flour is the excess reagent because you have enough to make 5 batches but you can only make 3 because of your amount of eggs. So you will have an excess amount of flour enough to make 2 more batches.
Method 1: It is all about the mole ratio.
Example using chemical equation: C3H8 + 5O2 ---> 3CO2 + 4H2O
a) Limiting reagent? Start with 14.8g C3H8 and 3.44g O2
Basic Example
cookie recipe:
2 eggs+ 2 cups of flour ------> 1 batch
At home I have 6 eggs and 10 cups of flour. How many batches of cookies can I make?
6 eggs x 1 batch/ 2eggs = 3 batches (maximum you can make)
10 cups of flour x 1 batch/2 cups of flour = 5batches
The eggs are the limiting reagent because whether you have 10 cups of flour of 100 cups of flour you only have enough eggs to make 3 batches. The cups of flour is the excess reagent because you have enough to make 5 batches but you can only make 3 because of your amount of eggs. So you will have an excess amount of flour enough to make 2 more batches.
Method 1: It is all about the mole ratio.
Example using chemical equation: C3H8 + 5O2 ---> 3CO2 + 4H2O
a) Limiting reagent? Start with 14.8g C3H8 and 3.44g O2
If I reacted all 14.8g C3H8, I would need 53.8g O2.
Why is O2 the limiting reagent?
- Because all 14.8g of C3H8 would require 53.8g O2 and we only have 3.44g of O2.
b) What is the amount of excess left over?
Why is O2 the limiting reagent?
- Because all 14.8g of C3H8 would require 53.8g O2 and we only have 3.44g of O2.
b) What is the amount of excess left over?
Method 2
Using the same chemical equation...
Using the same chemical equation...