Football
You're in the park with a few of your friends and you want to play a game of football. You decide that the most fair way to split up the teams is to have an even number of boys and girls on each side. There are ten boys, so there can be five of them on each team. But here's the problem, There are only nine girls! In this situation, the limiting reagent is the number of girls because there aren't enough girls to have an even second team.
This is how limiting reagents work. When you're making a chemical compound, you need to have a certain number of each reactant. If there isn't enough of one reactant to make the desired compound, then this element is the limiting reagent because it limits the number of compounds that can be made.
This is how limiting reagents work. When you're making a chemical compound, you need to have a certain number of each reactant. If there isn't enough of one reactant to make the desired compound, then this element is the limiting reagent because it limits the number of compounds that can be made.
Finding the Limiting Reagent
You may be thinking that finding limiting reagents is easy, and it is, but finding real limiting reagents is a little harder than splitting up a couple of football teams. So here's how you do it!
Step 1: Find the GFW of each reactant by finding its atomic mass.
Step 2: Divide the number of grams of each reactant by the GFW.
Step 3: Multiply each of the results by the coefficient of the other reactant.
Step 4: The compound with the lower number at the end of this process is the limiting reagent.
Step 1: Find the GFW of each reactant by finding its atomic mass.
Step 2: Divide the number of grams of each reactant by the GFW.
Step 3: Multiply each of the results by the coefficient of the other reactant.
Step 4: The compound with the lower number at the end of this process is the limiting reagent.
Examples:
1: Finding the limiting reagent if 37.5 grams of C6H6 were reacted with 63.8 grams of O2.
2C6H6 + 15O2 --> 12CO + 6H2O
First we find the GFW of each reactant.
GFW of C6H6: 78
GFW of O2: 32
Next we divide the number of grams of each reactant by the GFW.
37.5/78= 0.48
63.8/32= 1.99
Finally, we multiply each of the results by the coefficient of the other reactant/
0.48 x 15= 7.2
1.99 x 2= 3.99
Because 3.99 is smaller than 7.2, we know that the limiting reagent is O2.
2:Finding the limiting reagent if 31.3 grams of Na were reacted with 36.0 grams o fH2O.
2Na + 2H2O --> 2NaOH + H2
First we find the GFW of each reactant.
GFW of Na: 23
GFW of H2O: 32
Next we divide the number of grams of each reactant by the GFW.
31.3/23= 1.36
36/18= 2.0
Finally, we multiply each of the results by the coefficient of the other reactant/
1.36 x 2= 2.72
2.0 x 2= 4.0
Because 2.72 is smaller than 4.0, we know that the limiting reagent is O2.
2C6H6 + 15O2 --> 12CO + 6H2O
First we find the GFW of each reactant.
GFW of C6H6: 78
GFW of O2: 32
Next we divide the number of grams of each reactant by the GFW.
37.5/78= 0.48
63.8/32= 1.99
Finally, we multiply each of the results by the coefficient of the other reactant/
0.48 x 15= 7.2
1.99 x 2= 3.99
Because 3.99 is smaller than 7.2, we know that the limiting reagent is O2.
2:Finding the limiting reagent if 31.3 grams of Na were reacted with 36.0 grams o fH2O.
2Na + 2H2O --> 2NaOH + H2
First we find the GFW of each reactant.
GFW of Na: 23
GFW of H2O: 32
Next we divide the number of grams of each reactant by the GFW.
31.3/23= 1.36
36/18= 2.0
Finally, we multiply each of the results by the coefficient of the other reactant/
1.36 x 2= 2.72
2.0 x 2= 4.0
Because 2.72 is smaller than 4.0, we know that the limiting reagent is O2.
Quiz
Now test your skills by completing the problems below.
Question 1: Find the limiting reagent if 85.6 grams of Al were reacted with 498 grams of MnO.
2Al+3MnO=>Al2O3+3Mn
A: Al
B: MnO
2Al+3MnO=>Al2O3+3Mn
A: Al
B: MnO
Question 2: Find the limiting reagent if 39.9 grams of SiO2 were reacted with 77.4 grams of H2F2.
SiO2+2H2F2=>SiF4+2H2O
A: SiO2
B: H2F2
SiO2+2H2F2=>SiF4+2H2O
A: SiO2
B: H2F2
Question 3: Find the limiting reagent if 44.2 grams of P were reacted with 622 grams of Br2
2P+3Br2=>2PBr3
A: P
B: Br2
2P+3Br2=>2PBr3
A: P
B: Br2
Question 4: Find the limiting reagent if 98.7 grams of C4H10S2 were reacted with 386 grams of O2.
2C4H10S2+19O2=>8CO2+10H2O+4SO3
A: C4H10S2
B:O2
2C4H10S2+19O2=>8CO2+10H2O+4SO3
A: C4H10S2
B:O2
Question 5: Find the limiting reagent if 47.3 grams of PCl3 were reacted with 14.6 grams of H2O.
PCl3+3H2O=>H3PO3+3HCl
A:PCl3
B:H2O
PCl3+3H2O=>H3PO3+3HCl
A:PCl3
B:H2O
Question 6: Find the limiting reagent if 89.7 grams of C2H3O2F were reacted with 16.0 grams of O2.
4C2H3O2F+3O2=>8CO+6H2O+2F2
A:C2H3O2F
B:O2
4C2H3O2F+3O2=>8CO+6H2O+2F2
A:C2H3O2F
B:O2
Question 7: Find the limiting reagent if 50.9 grams of CH2F2 were reacted with 73.9 grams of O2?
2CH2F2+3O2=>2CO2+2H2O+2F2
A:CH2F2
B:O2
2CH2F2+3O2=>2CO2+2H2O+2F2
A:CH2F2
B:O2
Question 8: Find the limiting reagent if 99.6 grams of NH3 were reacted with 473 grams of O2.
4NH3+5O2=>4NO+6H2O
A:NH3
B:O2
4NH3+5O2=>4NO+6H2O
A:NH3
B:O2
Question 9: Find the limiting reagent if 7.38 grams of C2H4 were reacted with 15.2 grams of O2.
4NH3+5O2=>4NO+6H2O
A:C2H4
B:O2
4NH3+5O2=>4NO+6H2O
A:C2H4
B:O2
Question 10: Find the limiting reagent if 6.59 grams of C2H3OF were reacted with 3.40 grams of O2.
4C2H3OF+5O2=>8O+6H2O+2F2
A:C2H3OF
B:O2
4C2H3OF+5O2=>8O+6H2O+2F2
A:C2H3OF
B:O2
Cookies
Now let's imagine you're making a batch of chocolate chip cookies with exactly four chocolate chips in each cookie. You have exactly enough cookie dough to make ten cookies, and you have thirty chocolate chips.
You should now be capable of knowing that the limiting factor in the scenario is the chocolate chips, but we still don't know how to use this information. The next step is to figure out how many cookies can be made. It shouldn't be hard to tell that in this situation, you can make eight cookies.This number is called the theoretical yield. Although finding the theoretical yield was easy in this example, it's a little bit harder when it comes to chemistry.
You should now be capable of knowing that the limiting factor in the scenario is the chocolate chips, but we still don't know how to use this information. The next step is to figure out how many cookies can be made. It shouldn't be hard to tell that in this situation, you can make eight cookies.This number is called the theoretical yield. Although finding the theoretical yield was easy in this example, it's a little bit harder when it comes to chemistry.
Using the Limiting Reagent
(Theoretical Yield)
In this section, I'll explain how to use a limiting reagent to find theoretical yield. Let's get to it!
Step 1: Find limiting reagent using the skills learned in the first section.
Step 2: State the coefficient of the limiting reagent and the product under consideration as a ratio.
Step 3: Divide both coefficients by the coefficient of the limiting reagent.
(First part of the ratio will always be one)
Step 4: Multiply the 2nd number in the ratio by the number of moles in the limiting reagent.
Step 5: Multiply the answer from step 4 by the GFW of the product.
Step 1: Find limiting reagent using the skills learned in the first section.
Step 2: State the coefficient of the limiting reagent and the product under consideration as a ratio.
Step 3: Divide both coefficients by the coefficient of the limiting reagent.
(First part of the ratio will always be one)
Step 4: Multiply the 2nd number in the ratio by the number of moles in the limiting reagent.
Step 5: Multiply the answer from step 4 by the GFW of the product.
Examples:
1: How many grams of O2 would be produced if 95.7 grams of CO2 were reacted with 31.3 grams of H2O?
3CO2+4H2O --> C2H85O2
First we find the limiting reagent.
CO2
Next we put the coefficient of the limiting reagent and the product into a ratio.
4:5
Next we divide this by the coefficient of the limiting reagent.
4/4=1 : 5/4=1.25
We then multiply the second number in the ratio by the number of moles in the limiting reagent.
1.25 x 1.738= 2.17
Finally, we multiply this result by the GFW of the product
2.17 x 32= 67.34 grams
That final number is your answer.
3CO2+4H2O --> C2H85O2
First we find the limiting reagent.
CO2
Next we put the coefficient of the limiting reagent and the product into a ratio.
4:5
Next we divide this by the coefficient of the limiting reagent.
4/4=1 : 5/4=1.25
We then multiply the second number in the ratio by the number of moles in the limiting reagent.
1.25 x 1.738= 2.17
Finally, we multiply this result by the GFW of the product
2.17 x 32= 67.34 grams
That final number is your answer.
Quiz
Now test your skills by completing the problems below.
Question 1: How many grams of CO2 would be produced if 64.2 grams of C4H10O were reacted with 290 grams of O2?
C4H10O+6O2-->4CO2+5H2O
C4H10O+6O2-->4CO2+5H2O
Question 2: How many grams of Br2 would be produced if 62.4 grams of CH2Br2 were reacted with 22.2 grams of O2?
2CH2Br2+3O2-->2CO2+2H2O+2Br2
2CH2Br2+3O2-->2CO2+2H2O+2Br2
Question 3: How many grams of CO2 would be produced if 2.58 grams of C2H3OCl were reacted with 1.16 grams of CO2?
C2H3OCl+2O2-->2CO2+H2O+HCl
C2H3OCl+2O2-->2CO2+H2O+HCl
Question 4: How many grams of O2 would be produced if 93.7 grams of Na2O2 were reacted with 16.4 grams of H2O?
2Na2O2+2H2O-->4NaOH+O2
2Na2O2+2H2O-->4NaOH+O2
Question 5: How many grams of H2O would be produced if 27.3 grams of C7H16 were reacted with 138 grams of O2?
C7H16+11O2-->7CO2+8H2O
C7H16+11O2-->7CO2+8H2O
Question 6: How many grams of H2O would be produced if 59.3 grams of N2H4 were reacted with 66.7 grams of H2O2?
N2H4+2H2O2-->N24H2O
N2H4+2H2O2-->N24H2O
Question 7: How many grams of H2O would be produced if 31.5 grams of C2H5NSCl were reacted with 72.8 grams of O2?
C2H5NSCl+5O2-->2CO2+2H2O+2NO+SO3+HCl
C2H5NSCl+5O2-->2CO2+2H2O+2NO+SO3+HCl
Question 8: How many grams of SO3 would be produced if 85.7 grams of C4H8S2 were reacted with 297 grams of O2?
C4H8S2+9O2-->4CO2+4H2O+2SO3
C4H8S2+9O2-->4CO2+4H2O+2SO3
Question 9: How many grams of H2O would be produced if 42.5 grams of C2H3Cl3 were reacted with 11.6 grams of O2?
4C2H4Cl3+7O2-->4CO+6H2O+6Cl2
4C2H4Cl3+7O2-->4CO+6H2O+6Cl2
Question 10: How many grams of H2O would be produced if 78.8 grams of C2H3Cl3 were reacted with 34.1 grams of O2?
4C2H4Cl3+11O2-->8CO2+6H2O+6Cl2
4C2H4Cl3+11O2-->8CO2+6H2O+6Cl2
Cookie Reprise
Yes, that's right, we're going back to the cookie example again. So let's say you've made all of you cookies and you've got ten beautiful chocolate chip cookies. But wait! One of those glorious cookies is burnt and completely uneatable. You expected to yield 10 cookies, but you only got nine. Therefore, your percentage yield is 90%. Well guess what, this relates to science. When you calculate the theoretical yield, what you are actually calculating is the number of compounds you will produce in an ideal world. But hey, this is the real world and mistakes happen. So the percentage yield is essentially a number that tells us how close we got to the theoretical yield.
Percentage Yield
So we sortof know what percentage yield is from the last section, but like the other sections, it's a little more complex. So let's figure out how to do it.
Step 1: Evaluate the equation to determine what you are solving for (you may be using the theoretical yield to find the percentage yield or the other way around).
Step 2: Find the theoretical yield using skills we used from the last section.
Step 3: Use the equation (Actual Yield)/(Theoretical Yield) x 100.
Step 4: The result is your answer.
Step 2: Find the theoretical yield using skills we used from the last section.
Step 3: Use the equation (Actual Yield)/(Theoretical Yield) x 100.
Step 4: The result is your answer.
Examples
How many grams of H2O would be produced if the reaction of 32.1 grams of O2 produces a 72.0% yield?
C6H6O+4O2=>6CO+3H2O
we first find the theoretical yield.
13.536
We then multiply this number by the percentage yield/100
13.536 x (72.0/100)= 9.74 mol H2O
That result is our answer.
C6H6O+4O2=>6CO+3H2O
we first find the theoretical yield.
13.536
We then multiply this number by the percentage yield/100
13.536 x (72.0/100)= 9.74 mol H2O
That result is our answer.
Quiz
Question 1: How many grams of C would be produced if the reaction of 16.0 grams of N2 produces a 84.8% yield?
CaC2+N2=>Ca+CN2+C
CaC2+N2=>Ca+CN2+C
Question 2: How many grams of CO2 would be produced if the reaction of 1.21 grams of O2 produces a 62.0% yield?
2C2H3O2Br+3O2=>4CO2+2H2O+2HBr
2C2H3O2Br+3O2=>4CO2+2H2O+2HBr
Question 3: How many grams of CH4 would be produced if the reaction of 62.2 grams of CH3CHO produces a 47.8% yield?
CH3CHO-->CH4+CO2
CH3CHO-->CH4+CO2
Question 4: How many grams of CO2 would be produced if the reaction of 17.1 grams of C4H10O produces a 35.1% yield?
C4H10O+6O2-->4CO2+5H2O-->CH4+CO2
C4H10O+6O2-->4CO2+5H2O-->CH4+CO2
Question 5: What is the percent yield if the reaction of 73.8 grams of O2 produces 25.6 grams of H2O?
2C3H8O2S+11O2=>6CO2+8H2O+2SO3
2C3H8O2S+11O2=>6CO2+8H2O+2SO3
Finally...
Congratulations! You have reached the final test on this site. My goal has been to teach you how to find, use, and make sense of the limiting reagents. In this final section, we will combine all of this information into one big problem and see if I have taught you well.
First let's put this all into one big real life example one last time. We will use the cookies. Let's say you want to make the famous chocolate chip cookies with exactly four chocolate chips in each cookie. Again, you have exactly enough cookie dough to make ten cookies, and you have thirty chocolate chips. We know that the limiting factor here is the chocolate chips. We also know that you can make exactly eight cookies. Now let's say that again you burn one of those eight glorious cookies. You percentage yield is 87.5%. Now let's see if you can do it with elements.
First let's put this all into one big real life example one last time. We will use the cookies. Let's say you want to make the famous chocolate chip cookies with exactly four chocolate chips in each cookie. Again, you have exactly enough cookie dough to make ten cookies, and you have thirty chocolate chips. We know that the limiting factor here is the chocolate chips. We also know that you can make exactly eight cookies. Now let's say that again you burn one of those eight glorious cookies. You percentage yield is 87.5%. Now let's see if you can do it with elements.
The Ultimate Test
If 2CH2Cl2+3O2=>2CO2+2H2O+2Cl2, if 13.0 grams of CH2Cl2 were reacted with 48.6 grams of O2, first find the limiting reagent, then the theoretical yield of Cl2 and knowing that 32 is the actual yield, find the percentage yield.
Congratulations!!!
You have finally reached the end of this site!!! To continue practice, you can redo the above quizzes or go to the "more practice" section in the upper left corner. Thank you for learning about elements with me!