1. Chapter 11 Stoichiometry
The study of quantitative, or measurable, relationships that exist in chemical formulas and chemical reactions.

2. Mole – Mole Relationship
Use a balanced chemical equation to determine the mole – mole relationship of any substance.
Mole – Mole relationship can be applied to any substance in the balanced equation.
N2H H2O2  N2 + 4 H2O
According to this equation:

3. Practice Mole-Mole
Determine the number mole-mole relationship for Ag in the following equation:
3 AgNO3 + Al  Al(NO3)3 + 3 Ag

4. Mole – Mole Problems 1 mol Zn = 2 mol HCl
How many moles of HCl are needed to react with 2.3 moles of Zn?
2 HCl + Zn  ZnCl2 + H2
1 mol Zn = 2 mol HCl

5. Practice Problems
1) How many moles of Na are needed to react with 55.5 moles of Cl2, in the production of salt?
Na + Cl2  NaCl
2) How many moles of Al(OH)3 are produced when 2.53 moles of H2O react.
Al2O3 + H2O  Al(OH)3

6. 11-2 Solving Stoichiometry Problems
Using all other conversions from chapter 10, along with the mole-mole to determine unknown amounts of each substance.
1 mol = 6.02 x 1023 particles
1 mol = 22.4L at STP
1 mol = formula mass(g)

7. Example: Mass - Mass
Determine the mass of sodium hydroxide produced when .25g of sodium reacts with water.
2 Na + 2H2O  2NaOH + H2
.25g x g
2mol Na = 2 mol NaOH

8. Practice Problem #1
What mass of bromine, is produced when fluorine reacts with 3.55 moles of potassium bromide?
F2 + 2 KBr  2 KF + Br2
3.55 mol x g

9. Mass – Volume : Example
Find the mass of aluminum required to produce 1.32 L of hydrogen gas at STP.
2 Al + 3 H2SO4  Al2(SO4)3 + 3 H2
x g L

10. Volume – Volume: Relationship
The volume relationship of 2 gases at STP are directly proportional to the mole-mole relationship of those 2 gases.
Why?
All gases at STP occupy the same volume, 22.4L / mol.

11. Example: Volume - Volume
What volume of hydrogen gas is necessary to produce 16.0L of ammonia (NH3)?
N H2  2 NH3
x L L

12. Chapter 11–3: Limiting and Excess Reactants
Limiting reactant
The reactant that limits the amount of product obtained.
Completely consumed in the reaction. No trace remains in the product.
Excess reactant
The reactant that isn’t completely used in the reaction and remains after the reaction is over.

13. Example Problem: 1st determine the limiting reactant.
Determine the mass of NaCl produced when 1.2 mol of Na react with .85 mol Cl2.
2 Na + Cl2  2 NaCl
1.2mol mol ?mol
1st determine the limiting reactant.
Convert moles of one given to moles of another.

14. Determining Limiting and Excess
Compare: Have vs. Need
Have = Given amount
Need = Calculated amount
If Have > Need, excess reactant
If Need > Have, limiting reactant

15. Comparing You have .85mol Cl2 you need .6mol Cl2.
Determine the mass of NaCl produced when 1.2 mol of Na react with .85 mol Cl2.
2 Na + Cl2  2 NaCl
1.2mol mol ?g
You have .85mol Cl2 you need .6mol Cl2.
Chlorine is the excess, which means Sodium is the limiting.

16. Finishing the Problem
Determine the mass of NaCl produced when 1.2 mol of Na react with .85 mol Cl2.
2 Na + Cl2  2 NaCl
1.2mol mol ?g
You must begin with the limiting given to determine the unknown amount.

17. Practice Problem 3ZnO + 2Al(NO3)3  Al2O3 + 3Zn(NO3)2
Considering 25g of ZnO react with .555moles of Al(NO3)3. How many grams of Al2O3 will be produced?
Determine limiting/excess reactant.
Determine excess that will remain.
Determine the unknown.

18. Percent Yield Yield = Product Actual Yield
Amount of product given in the problem.
Theoretical Yield (Expected)
Amount of product determined by a stoichiometry calculation.
Percent Yield

19. Sample Problem
Determine the percent yield when 2.80g Al(NO3)3 react to produce .966g of Al(OH)3.
Al(NO3)3 + 3NaOH  Al(OH)3 + 3NaNO3
2.80g g

20. Identifying what to do! 1) Regular stoichiometry: 2) Limiting/Excess:
Given one value.
2) Limiting/Excess:
Given 2 values both are reactants (left side of equation)
3) %yield:
Given 2 values 1 reactant and 1 product
All problems involve a mole-mole conversion.

21. Practice Problem
Determine the percent yield when 1.50g Al(OH)3 decomposes to produce .25g of H2O.
2Al(OH)3  Al2O3 + 3H2O
1.50g g