How to Use Limiting Reagents to Find the Theoretical Yield
Using Limiting Reagents requires all of the skills you used for finding limiting reagents, but taking it to the next level.
Everyday Examples:
In the examples below you will be finding the theoretical yield of one of the compounds produced in the reaction. The same method can be used to find the theoretical yield of each of the other compounds that are produced.
Example problem:
Step 1: Find the limiting reagent:
For the balanced equation shown below, if 96.1 grams of C2H3F were reacted with 142 grams of O2, how many grams of CO would be produced?
2C2H3F+3O2=>4CO+2H2O+2HF
C2H3F: 96.1 grams / 46 grams/mole = 2.089 moles (3) = 6.267391304347827 moles
O2: 142 grams / 32 grams/mole = 4.4375 moles (2) = 8.875 moles
6.267 moles < 8.875 moles
C2H3F is the limiting reagent.
Step 2: Compare the coefficients:
In the equation, look at the coefficient of the limiting reagent and of the compound for which you want to find the number of grams that will be produced by the reaction. They are expressed as:
2C2H3F and 4CO
Compare the coefficients through a ratio.
2 : 4 (simplified to) --> 1 : 2
If you have a much more complicated ratio, such as 13 : 4 , then you can reduce the ratio through decimals. 13/13 = 1, 4/13 = 0.308
To find the number of moles for CO, multiply the decimal you got from the ratio by the original number of moles from C2H3F: (0.308)(2.089) = 0.643 moles (for CO).
Step 3: Find the number of grams for the desired compound:
Now we have already found how many moles the limiting reagent has. To find the amount of moles for the desired compound, use the ratio. Once you have found the amount of moles for the desired compound, use the equation: Moles = Grams / GFW
Multiply the amount of moles by the GFW of the compound to find the amount of grams.
2.089 moles (2) = 4.178 moles
4.178 (28) = 117 grams
Here are two more example problems:
For the balanced equation shown below, if 23.1 grams of C2H3O2Cl were reacted with 2.26 grams of O2, how many grams of H2O would be produced?
2C2H3O2Cl+O2=>4CO+2H2O+2HCl
Step 1:
C2H3O2Cl: 23.1 grams / 94.5 grams/mole = 0.25 moles
O2: 2.26 grams / 32 grams/mole = 0.071 moles (2) = 0.141 moles
O2 is the limiting reagent.
Step 2:
1 : 2
2/1 = 2
Step 3:
0.071 moles (2) = 0.141 moles (18) = 2.543 grams
This question is more difficult, because of the difference in ratios:
For the balanced equation shown below, if 44.4 grams of C4H8S were reacted with 88.9 grams of O2, how many grams of CO2 would be produced?
2C4H8S+15O2=>8CO2+8H2O+2SO3
C4H8S: 44.4 grams / 88 grams/mole = 0.505 mol (15) = 7.568 moles
O2: 88.9 grams / 32 grams/mole = 2.778125 moles (2) = 5.556 moles
O2 is the limiting reagent.
15 : 8
44.4 (8) = 325.08
32 (15) = 480
352.08/480 = 0.7335
0.7335 (88.9) = 65.208 grams
Ready to move onto the quiz?
Everyday Examples:
- Here you are again with your sushi restaurant. You have 20 slices each of eel and avocado. It takes 2 slices of eel and 1 slice of avocado to make an eel avocado roll. Each order of sushi comes with 6 rolls. The theoretical yield would be 10 eel avocado rolls, because eel is your limiting reagent, and it will run out before you can finish the second order of sushi.
- Business is booming on the high seas. You catch 12 salmon and 12 tuna. The sushi restaurant needs to make 24 salmon tuna rolls. Each salmon yields enough for 2 rolls, and each tuna yields enough for 1 roll. The sushi restaurant will only be able to make 12/24 of the sushi rolls it needs, making the theoretical yield 12 rolls.
In the examples below you will be finding the theoretical yield of one of the compounds produced in the reaction. The same method can be used to find the theoretical yield of each of the other compounds that are produced.
Example problem:
Step 1: Find the limiting reagent:
For the balanced equation shown below, if 96.1 grams of C2H3F were reacted with 142 grams of O2, how many grams of CO would be produced?
2C2H3F+3O2=>4CO+2H2O+2HF
C2H3F: 96.1 grams / 46 grams/mole = 2.089 moles (3) = 6.267391304347827 moles
O2: 142 grams / 32 grams/mole = 4.4375 moles (2) = 8.875 moles
6.267 moles < 8.875 moles
C2H3F is the limiting reagent.
Step 2: Compare the coefficients:
In the equation, look at the coefficient of the limiting reagent and of the compound for which you want to find the number of grams that will be produced by the reaction. They are expressed as:
2C2H3F and 4CO
Compare the coefficients through a ratio.
2 : 4 (simplified to) --> 1 : 2
If you have a much more complicated ratio, such as 13 : 4 , then you can reduce the ratio through decimals. 13/13 = 1, 4/13 = 0.308
To find the number of moles for CO, multiply the decimal you got from the ratio by the original number of moles from C2H3F: (0.308)(2.089) = 0.643 moles (for CO).
Step 3: Find the number of grams for the desired compound:
Now we have already found how many moles the limiting reagent has. To find the amount of moles for the desired compound, use the ratio. Once you have found the amount of moles for the desired compound, use the equation: Moles = Grams / GFW
Multiply the amount of moles by the GFW of the compound to find the amount of grams.
2.089 moles (2) = 4.178 moles
4.178 (28) = 117 grams
Here are two more example problems:
For the balanced equation shown below, if 23.1 grams of C2H3O2Cl were reacted with 2.26 grams of O2, how many grams of H2O would be produced?
2C2H3O2Cl+O2=>4CO+2H2O+2HCl
Step 1:
C2H3O2Cl: 23.1 grams / 94.5 grams/mole = 0.25 moles
O2: 2.26 grams / 32 grams/mole = 0.071 moles (2) = 0.141 moles
O2 is the limiting reagent.
Step 2:
1 : 2
2/1 = 2
Step 3:
0.071 moles (2) = 0.141 moles (18) = 2.543 grams
This question is more difficult, because of the difference in ratios:
For the balanced equation shown below, if 44.4 grams of C4H8S were reacted with 88.9 grams of O2, how many grams of CO2 would be produced?
2C4H8S+15O2=>8CO2+8H2O+2SO3
C4H8S: 44.4 grams / 88 grams/mole = 0.505 mol (15) = 7.568 moles
O2: 88.9 grams / 32 grams/mole = 2.778125 moles (2) = 5.556 moles
O2 is the limiting reagent.
15 : 8
44.4 (8) = 325.08
32 (15) = 480
352.08/480 = 0.7335
0.7335 (88.9) = 65.208 grams
Ready to move onto the quiz?