1. Limiting Reagents
Problem: Find the amount of cakes we can make when we have 4 sticks of butter and 50 cups of flour.
1 Butter + 2 Flour -> 1 Cake
We need to find out which one of or reactants is our limiting reagent.
So we do some calculations.
If I have 4 sticks of butter, then I use up 8 cups of flour.
Since I have 50 cups of flour, and I only used 8, then I have 42 cups of flour left over in excess reagent. Since flour is my excess, this means butter must be my limiting reagent.
Now that I have found out that butter is my limiting reagent I will solve the problem with that knowledge. I have 4 sticks of butter, so I can only produce 4 cakes!

2. Limiting Reagents
Determine the mass of P4O10 formed if 25g P4 and 50g O2 are combined.
P4 + 5O2  P4O10
What is the given? We have to find the limiting reagent!
1 Mole P Mole O g O2
25 g P4 X X X = g O2
124 g P Mole P Mole O2

3. Limiting Reagent
32.26 O2 was what we got when we used up all of our P4.
We have to now compare how much we used, to how much we had.
We had 50 g O2.
We used g O2.
We therefore had g O2 left over, in other words, in excess.
Since we have O2 left over, but have used up all of our P4, we now know that O2 is our excess reagent.
This means that P4 is our limiting reagent.

4. Limiting Reagent
We now know to use P4, our limiting reagent as our given.
This is because we can only produce as much product as our limiting reagent will allow.
Think of it this way: We have 5 bicycle frames, and 20 wheels, how many bicycles can we produce?
Frames: 5 bikes
Wheels: 10 bikes
We can only make 5 bikes, because frames are our limiting reagent!

5. Limiting Reagents
Determine the mass of P4O10 formed if 25g P4 and 50g O2 are combined.
P4 + 5O2  P4O10
The limiting reagent was P4, therefore we use P4 as our given to solve.
1 Mole P Mole P4O g P4O10
25 g P4 X X X = g P4O10
124 g P Mole P Mole P4O10

6. Percent Yield Calculations
If 100 g of Na and 100 g of Fe2O3 are used in this reaction, how much Na2O, can be made?
6Na + Fe2O3  Na2O + 2Fe
1 Mole Na Mole Fe2O g Fe2O3
100 g Na X X X = g Fe2O3
23 g Na Mole Na Mole Fe2O Fe2O3 is limiting

7. Percent Yield Calculations
If 100 g of Fe2O3 and an excess of Na are used in this reaction, how much Na2O, can be made?
6Na + Fe2O3  Na2O + 2Fe
1 Mole Fe2O Mole Na2O g Na2O
100 g Fe2O3 X X X = g Na2O
160 g Fe2O Mole Fe2O Mole Na2O This is the theoretical yield!

8. Percent Yield Calculations
Calculate the percent yield if the reaction actually yields 28.6 g of Na2O.
38.75 g Na2O was our theoretical. It is our theoretical because we had to do the math on paper.
Actual
Theoretical
X 100 =
X 100 = 73.81%

9. Using Volume With Stoichiometry
What volume of NH3 at STP is produced if 30.0 g of N2 is reacted with an excess of H2?
N2+3H2→ 2NH3

10. Using Volume With Stoichiometry
How many liters of propane gas (C3H8) are needed to make 75g of water?
C3H8+ 5O2 -> 3CO2+ 4H2O