Limiting Reagents
Because life rarely gives you exactly the right amount of stuff
Limiting Reagents
At this point, you will have done the S'mores activity in class, hopefully.
Given that, you should have some idea that because things can be mixed
in whatever proportion you want, but only react in a certain proportion,
you will often end up running out of one reactant before the other.
Limiting Reagents
Therefore, this presentation is just about some techniques you can use.
Identifying the Limiting Reagent
Let's start with a sample reaction:
2Na3PO4 + 3CaO → Ca3(PO4)2 + 3Na2O
10 moles 12 moles
Identifying the Limiting Reagent
Let's start with a sample reaction:
2Na3PO4 + 3CaO → Ca3(PO4)2 + 3Na2O
10 moles 12 moles
Method 1:
Just calculate the yield using each compound normally:
10 moles Na3PO4
12 moles CaO
yields 5 moles Ca3(PO4)2 and 15 moles Na2O
yields 4 moles Ca3(PO4)2 and 12 moles Na2O
Identifying the Limiting Reagent
Let's start with a sample reaction:
2Na3PO4 + 3CaO → Ca3(PO4)2 + 3Na2O
10 moles 12 moles
Method 1:
Just calculate the yield using each compound normally:
10 moles Na3PO4
12 moles CaO
yields 5 moles Ca3(PO4)2 and 15 moles Na2O
yields 4 moles Ca3(PO4)2 and 12 moles Na2O
Whichever one gives you less is the one that will run out first (limiting)
Identifying the Limiting Reagent
Let's start with a sample reaction:
2Na3PO4 + 3CaO → Ca3(PO4)2 + 3Na2O
10 moles 12 moles
Method 1:
How much of the Na3PO4 will be used up? Use the ratio
8 moles of Na3PO4 will react with the 12 moles of CaO
Identifying the Limiting Reagent
Let's start with a sample reaction:
2Na3PO4 + 3CaO → Ca3(PO4)2 + 3Na2O
10 moles 12 moles
Method 2:
Figure out how many times you can do the reaction, based on the amount you
have, and how many it takes each time.
10 moles Na3PO4 / 2 = enough to do it 5 (moles of) times
12 moles CaO
/ 3 = enough to do it 4 (moles of) times
Identifying the Limiting Reagent
Let's start with a sample reaction:
2Na3PO4 + 3CaO → Ca3(PO4)2 + 3Na2O
10 moles 12 moles
Method 2:
Take this number of times, and multiply it by the coefficient of anything you're
interested in:
4 (moles of) times * 2 = 8 moles Na3PO4 used
4 (moles of) times * 1 = 4 moles of Ca3(PO4)2
4 (moles of) times * 3 = 12 moles of Na2O
Either Way We Get This
Let's start with a sample reaction:
2Na3PO4 + 3CaO → Ca3(PO4)2 + 3Na2O
Starting amount:
10 moles 12 moles
0
0
Amount Used:
8 moles 12 moles
Final Amount:
2 moles
0
4 moles
12 moles
Or, In Grams
Let's start with a sample reaction:
2Na3PO4 + 3CaO → Ca3(PO4)2 + 3Na2O
Starting amount:
1640g
673g
0
0
Total: 2313g
Final Amount:
Total: 2313g
328g
0
1241g
744g
Conservation of mass still applies, but you get less products out than the mass
you put in, because some of it never reacted.
Summary
• Two methods. Both give the same results.
• Everything still reacts in the reaction equation
proportions.
• You will have less products than what you put in, but
conservation of mass still applies—some just isn't used.
• S'mores are delicious.