How to Solve Limiting Reagents
Here's a step by step way to solve Limiting Reagents!
Okay lets say this is your question:
For the balanced equation shown below, what would be the limiting reagent if 20.1 grams of C7H16 were reacted with 52.7 grams of O2?
C7H16+11O2=>7CO2+8H2O
Is it C7H16?
Or it it O2?
Okay!
First thing is, you can completely ignore the second half of the equation, you don't need it.
All you look at is the part that says C7H16+11O2.
So to start solving, you look at how many grams of each molecule there are. In this case there are 20.1 grams of C7H16 and 52.7 grams ofO2.
The next part you'll need is to find the GFW of each of the molecules.
If you don't know what the GFW is, or how to find it click here.
Here the GFW for C7H16 is 100, and the GFW for O2 is 32.
Now you can start the solving.
To make things easy, set up the list of numbers like this:
C7H16 11O2
20.1 52.7
100 32
To start solving, divide the number of grammes by the GFW.
C7H16 11O2 So in this case, for C7H16 you divide 20.1 by 100 and you get 0.201,
20.1 52.7 and for O2 you divide 52.7 by 32 and you get 1.647
100 32
=0.201 =1.647
Now you have the number of moles, but finally to get the limiting reagent, you have to cross multiply the coefficient by the number of moles.
C7H16 11O2 So in this case, you multiply 11 (from the O2) by 0.201, and get 2.211,
20.1 52.7 and since there is only 1 (from the C7H16) 1.647 stays the same.
100 32
0.201 1.647
=2.211 =1.647
Now that you look at the two numbers above, the smaller number is the limiting reagent.
C7H16 11O2 In this case 1.647 is the smaller of the two, so the answer is the limiting reagent is O2!
20.1 52.7
100 32
0.201 1.647
2.211 1.647
Lets go through one more example a little faster this time.
Here's your next question:
For the balanced equation shown below, what would be the limiting reagent if 18.2 grams of FeS2 were reacted with 19.2 grams of O2?
4FeS2+11O2=>2Fe2O2+8SO2
Is it C7H16?
Or it it O2?
Okay first you take the first half of the equation:
4FeS2+11O2
Then take the number of grams:
4FeS2 11O2
18.2 19.2
Then find the GFWs of FeS2 and O2 and place them in the list of numbers:
4FeS2 11O2
18.2 19.2
116 32
Then divide:
4FeS2 11O2
18.2 19.2
116 32
=0.157 =0.6
Then cross multiply:
4FeS2 11O2
18.2 19.2
116 32
0.157 0.6
=1.727 =2.4
Then find the limiting reagent:
4FeS2 11O2 In this case 1.727 is the smaller number so the answer is the limiting reagent is FeS2!
18.2 19.2
116 32
0.157 0.6
1.727 2.4
Make Sense? How about practicing yourself?
Click here for a short QUIZ!
Okay lets say this is your question:
For the balanced equation shown below, what would be the limiting reagent if 20.1 grams of C7H16 were reacted with 52.7 grams of O2?
C7H16+11O2=>7CO2+8H2O
Is it C7H16?
Or it it O2?
Okay!
First thing is, you can completely ignore the second half of the equation, you don't need it.
All you look at is the part that says C7H16+11O2.
So to start solving, you look at how many grams of each molecule there are. In this case there are 20.1 grams of C7H16 and 52.7 grams ofO2.
The next part you'll need is to find the GFW of each of the molecules.
If you don't know what the GFW is, or how to find it click here.
Here the GFW for C7H16 is 100, and the GFW for O2 is 32.
Now you can start the solving.
To make things easy, set up the list of numbers like this:
C7H16 11O2
20.1 52.7
100 32
To start solving, divide the number of grammes by the GFW.
C7H16 11O2 So in this case, for C7H16 you divide 20.1 by 100 and you get 0.201,
20.1 52.7 and for O2 you divide 52.7 by 32 and you get 1.647
100 32
=0.201 =1.647
Now you have the number of moles, but finally to get the limiting reagent, you have to cross multiply the coefficient by the number of moles.
C7H16 11O2 So in this case, you multiply 11 (from the O2) by 0.201, and get 2.211,
20.1 52.7 and since there is only 1 (from the C7H16) 1.647 stays the same.
100 32
0.201 1.647
=2.211 =1.647
Now that you look at the two numbers above, the smaller number is the limiting reagent.
C7H16 11O2 In this case 1.647 is the smaller of the two, so the answer is the limiting reagent is O2!
20.1 52.7
100 32
0.201 1.647
2.211 1.647
Lets go through one more example a little faster this time.
Here's your next question:
For the balanced equation shown below, what would be the limiting reagent if 18.2 grams of FeS2 were reacted with 19.2 grams of O2?
4FeS2+11O2=>2Fe2O2+8SO2
Is it C7H16?
Or it it O2?
Okay first you take the first half of the equation:
4FeS2+11O2
Then take the number of grams:
4FeS2 11O2
18.2 19.2
Then find the GFWs of FeS2 and O2 and place them in the list of numbers:
4FeS2 11O2
18.2 19.2
116 32
Then divide:
4FeS2 11O2
18.2 19.2
116 32
=0.157 =0.6
Then cross multiply:
4FeS2 11O2
18.2 19.2
116 32
0.157 0.6
=1.727 =2.4
Then find the limiting reagent:
4FeS2 11O2 In this case 1.727 is the smaller number so the answer is the limiting reagent is FeS2!
18.2 19.2
116 32
0.157 0.6
1.727 2.4
Make Sense? How about practicing yourself?
Click here for a short QUIZ!