1. Mole to Mole, G, L, & other Notes

2. Remember that you can get the following information from balanced equations:
Reactants
Products
Moles
MOLE RATIOS
Molar/Formula Mass (using the periodic table)

3. New and Improved Mole Map
A super bridge has been created that allows you to convert between different substances.
Before you could convert between MASS, VOLUME, PARTICLES and MOLES….
Now you can convert from SUBSTANCE to SUBSTANCE using mole ratios.

4. Representative particles of G 1 mole G 6.02 x 1023
Representative particles of W
6.02 x 1023
1 mole W
Mass of W
mass W
1 mole W
Mass of G
1 mole G
mass G
b mole W
a mole G
Mol G
Mol W
Volume of W stp)
22.4 L W
1 mole W
Volume of G stp)
1 mole G
22.4 L G

5. 2H20(l)  2H2(g) + O2(g)
How many moles of O2 would you produce from 27.0 g H20?
So..think we are starting with ____ of ____and we want to get to ____ of _____
Lets look at the chart!
27.0 g
H2O O2
moles

6. 1. How many moles of O2 would you produce from 27. 0 g H20
1. How many moles of O2 would you produce from 27.0 g H20 ? 2H20(l)  2H2(g) + O2(g)
27.0 g H20 x 1 mol H x 1 mol O2 = mol O g H mol H20
Molar mass of H20
MOLE RATIO

7. Write what you start with
Sn(s) + 2HF(g)  SnF2(s) + H2(g) How many liters of HF are needed to produce 9.40 L H2 at STP?
Write what you start with
Convert from L of H2 to moles H2 (1 mol = 22.4 L)
Convert from mole H2 to moles HF (use mole ratio)
Convert to liters of HF
9.40 L H2 x 1 mol H2 x 2 mol HF x 22.4 L HF = 18.8 L HF
L H mol H mol HF

8. Review Homework Questions?
Now Practice! Do the Stoichiometry Worksheet B!
You have 30 minutes then we will do Limiting Reagents

9. Limiting Reagents
If you are working at a car factory and you have 20 frames and 40 tires, how many cars can you make?
You have enough frames for 20 cars but only enough tires for 10 cars, so you can only make 10 cars.
So the TIRES are the limiting factor..

10. Limiting Reagents
Using the balanced equation you can determine the number of moles need to produce a specific amount of the product
You can also determine which reactant you will “run out” of first – the limiting reagent

11. Example: 2H2(g) + O2 (g)  2H2O(l)
Using the equation above, which reactant will be the limiting reactant if you have 4.0 g of H2 and 6.0 g of O2?

12. Covert to moles of H2 and O2 Compare moles using mole ratio
2H2(g) + O2 (g)  2H2O(l) Using the equation above, which reactant will be the limiting reactant if you have 4.0 g of H2 and 6.0 g of O2?
Covert to moles of H2 and O2
Compare moles using mole ratio
4.0 g H2 x 1 mole H2 = 2.0 mol H2
2.0 g H2
6.0 g O2 x 1 mole O2 = 0.19 mol O2
32.0 g O2

13. Now you practice Problems on worksheet & then…
Practice Problem on pg 313
Pg 321: #22

14. Percent Yield Percent yield = actual yield x 100 theoretical yield
Actual yield is measured in the lab
Theoretical yield is found using the balanced equation

15. Example: 2H2(g) + O2 (g)  2H2O(l)
Using the equation above, if you started with1.0 L of H2 in the lab how many moles of water would you expect to make?

16. 1.0 L H2 x 1 mol H2 x 2 mol H2O = 0.045 mol H2O 22.4 L H2 2 mol H2
2H2(g) + O2 (g)  2H2O(l) Using the equation above, if you started with 1.0 L of H2 in the lab how many moles of water would you expect to make?
1.0 L H2 x 1 mol H2 x 2 mol H2O = mol H2O L H mol H2
Lets go one step further….
Using the information calculated, what would be the percent yield if in the lab you measure mol H2O?

17. Percent yield = actual yield x 100 theoretical yield
Using the information calculated, what would be the percent yield if in the lab you measure mol H2O?
Percent yield = actual yield x 100
theoretical yield
0.030 mol H2O x 100 = 67%
0.045 mol H2O

18. What do I need to know? Mole ratio
New and improved mole map (conversions)
1 mole = 22.4 L
1 mole = molar mass (g)
1 mole = 6.02x1023 particles
Limiting reagents
Percent yield
UNITS!!! on everything