Using Limiting Reagents
Now you may know how to find the Limiting Reagent, but you may be wondering "why does it matter? Where is this information used?"
Knowing how to find the Limiting Reagent, and learning how to use it can help you find out how much product you will get.
This is also known as the Theoretical Yield, it is how much of a certain product you would (in theory) get.
Here's a step by step way to use Limiting Reagents!
Okay lets say this is your question:
For the balanced equation shown below, if 61.5 grams of NO were reacted with 42.0 grams of O2, how many grams of NO2 would be produced?
2NO+O2=>2NO2
Okay! So the first thing you need to do is find the Limiting Reagent.
If you don't know how to do this, you should look at the page called "How to Solve Limiting Reagents!"
Once you find the Limiting Reagent, in this case you should have two lists of numbers that look like this:
2NO 9O2
61.5g 42.0g
30 32
=2.05 =1.313
=2.05 =2.626
So in this case, the Limiting Reagent is NO, but that doesn't answer the question.
We want to find how many grams of NO2 would be produced.
So we're gonna look back at the original equation.
2NO+O2=>2NO2
We want the ratio of the number of molecules of the Limited Reagent, versus the number of molecules of the compound we're trying to find. By just looking at the equation you can get the ratio.
In this case, you see the number 2 in front of NO, and you see the number 2 in front of NO2, therefore your first ratio is 2:2. But you want how many molecules of NO2 there would be if there was 1 molecule of NO.
Basically, you do this:
2 : 2
1 : 1
TADA! Make sense?
So now you know for every 1 molecule of NO, there is 1 molecule of NO2.
This still doesn't answer the question, you want to know how many grams of NO2 there are.
This is where you'll need a certain equation that looks like:
mol=g/GFW or specifically # of moles=# of grams/Gram Formula Weight
If you remember from "How to Solve Limiting Reagents" to find the Limiting Reagent, you actually find the number of moles!
2NO 9O2
61.5g 42.0g
30 32
=2.05 =1.313 <- I mean this row of numbers!
=2.05 =2.626
So you can use the ratio to figure out how many moles there are of NO2:
2 : 2
1 : 1
2.05 : 2.05
Easy enough. So now you can plug this into the equation:
2.05=g/GFW
Now because you want to find the number of grams, you need to multiply the number of moles by the GFW.
If you don't know what the GFW is, or how to find it click here.
In this case the GFW is 46 so:
2.05x46=g
2.05x46=94.3g
So the answer is 94.3 grams!
Easy enough? Let's just try one more.
Here's the question:
For the balanced equation shown below, if 49.1 grams of C2H5N were reacted with 83.1 grams of O2, how many grams of NO2 would be produced?
4C2H5N+13O2=>8CO+10H20+4NO2
Okay! First find the Limiting Reagent.
You should have a list of numbers that look like this:
4C2H5N 13O2
49.1g 83.1g
43 32
=1.142 =2.597
=14.846 =10.388
So in this case O2 is the Limiting Reagent, but we want to find how many grams of NO2 will be produced.
So we are gonna make a ratio based on the original equation:
4C2H5N+13O2=>8CO+10H20+4NO2
So we use the number before the Limiting Reagent and the number before the compound we're trying to find.
In this case it's the 13 in front of O2, and the 4 in front of NO2.
So our ratio will look like this:
13 : 4
1 : 4/13
Now we find how many moles of NO2 there are for the number of moles of O2:
13 : 4
1 : 4/13
2.597 : 0.799 Remember, you use the number of moles of the limiting reagent from that list you made earlier.
Okay then you plug the number of moles of NO2 into the equation below:
mol=g/GFW
0.799xGFW=g
The GFW in this case is 46.
Plug that in and solve:
mol=g/GFW
0.799xGFW=g
0.799x46=g
0.799x46=36.754g
So the answer is 36.754 grams!
Make Sense? How about practicing yourself?
Click here for a short Quiz!
Knowing how to find the Limiting Reagent, and learning how to use it can help you find out how much product you will get.
This is also known as the Theoretical Yield, it is how much of a certain product you would (in theory) get.
Here's a step by step way to use Limiting Reagents!
Okay lets say this is your question:
For the balanced equation shown below, if 61.5 grams of NO were reacted with 42.0 grams of O2, how many grams of NO2 would be produced?
2NO+O2=>2NO2
Okay! So the first thing you need to do is find the Limiting Reagent.
If you don't know how to do this, you should look at the page called "How to Solve Limiting Reagents!"
Once you find the Limiting Reagent, in this case you should have two lists of numbers that look like this:
2NO 9O2
61.5g 42.0g
30 32
=2.05 =1.313
=2.05 =2.626
So in this case, the Limiting Reagent is NO, but that doesn't answer the question.
We want to find how many grams of NO2 would be produced.
So we're gonna look back at the original equation.
2NO+O2=>2NO2
We want the ratio of the number of molecules of the Limited Reagent, versus the number of molecules of the compound we're trying to find. By just looking at the equation you can get the ratio.
In this case, you see the number 2 in front of NO, and you see the number 2 in front of NO2, therefore your first ratio is 2:2. But you want how many molecules of NO2 there would be if there was 1 molecule of NO.
Basically, you do this:
2 : 2
1 : 1
TADA! Make sense?
So now you know for every 1 molecule of NO, there is 1 molecule of NO2.
This still doesn't answer the question, you want to know how many grams of NO2 there are.
This is where you'll need a certain equation that looks like:
mol=g/GFW or specifically # of moles=# of grams/Gram Formula Weight
If you remember from "How to Solve Limiting Reagents" to find the Limiting Reagent, you actually find the number of moles!
2NO 9O2
61.5g 42.0g
30 32
=2.05 =1.313 <- I mean this row of numbers!
=2.05 =2.626
So you can use the ratio to figure out how many moles there are of NO2:
2 : 2
1 : 1
2.05 : 2.05
Easy enough. So now you can plug this into the equation:
2.05=g/GFW
Now because you want to find the number of grams, you need to multiply the number of moles by the GFW.
If you don't know what the GFW is, or how to find it click here.
In this case the GFW is 46 so:
2.05x46=g
2.05x46=94.3g
So the answer is 94.3 grams!
Easy enough? Let's just try one more.
Here's the question:
For the balanced equation shown below, if 49.1 grams of C2H5N were reacted with 83.1 grams of O2, how many grams of NO2 would be produced?
4C2H5N+13O2=>8CO+10H20+4NO2
Okay! First find the Limiting Reagent.
You should have a list of numbers that look like this:
4C2H5N 13O2
49.1g 83.1g
43 32
=1.142 =2.597
=14.846 =10.388
So in this case O2 is the Limiting Reagent, but we want to find how many grams of NO2 will be produced.
So we are gonna make a ratio based on the original equation:
4C2H5N+13O2=>8CO+10H20+4NO2
So we use the number before the Limiting Reagent and the number before the compound we're trying to find.
In this case it's the 13 in front of O2, and the 4 in front of NO2.
So our ratio will look like this:
13 : 4
1 : 4/13
Now we find how many moles of NO2 there are for the number of moles of O2:
13 : 4
1 : 4/13
2.597 : 0.799 Remember, you use the number of moles of the limiting reagent from that list you made earlier.
Okay then you plug the number of moles of NO2 into the equation below:
mol=g/GFW
0.799xGFW=g
The GFW in this case is 46.
Plug that in and solve:
mol=g/GFW
0.799xGFW=g
0.799x46=g
0.799x46=36.754g
So the answer is 36.754 grams!
Make Sense? How about practicing yourself?
Click here for a short Quiz!