1. Stoichiometry L

3. What is it really? Cooking Method of finding
how many cookies you can make given these ingredients
how many ingredients you had to make this many cookies

4. All goes back to balanced equations
First write the equation and balance it.
You are observing the combustion of methane, CH4. Your teacher asks you how many moles of water are expected to form in the complete combustion of 2.50 mol of methane.
CH4 + O2  CO2 + H2O UNBALENCED
CH4 + 2O2  CO2 + 2H2O BALENCED

5. Now the math
Remember that the balanced equations shows the relationship of how many moles of each compound is used in this reaction
Like the ingredients on a recipe (3 eggs + ½ cup water = eggy water)
So the coefficient of one compound to the next is the mole ratio

6. Now the math (….continued)
You are observing the combustion of methane, CH4. Your teacher asks you how many moles of water are expected to form in the complete combustion of 2.50 mol of methane.
CH4 + 2O2  CO2 + 2H2O
So if we had 1 mole of methane we would produce 2 moles of water
But since we have 2.50 mol CH4…

7. The math
You are observing the combustion of methane, CH4. Your teacher asks you how many moles of water are expected to form in the complete combustion of 2.50 mol of methane.
CH4 + 2O2  CO2 + 2H2O
Therefore we would expect to retrieve 5 mols of H2O from the combustion of 2.5 mol of methane
Mole ratio
5 mol H2O

8. mole to mole
The synthesis reaction between nitrogen monoxide and oxygen forms nitrogen dioxide.
How many moles of nitrogen monoxide would be needed to fully react with 3.6 moles of oxygen
How many moles of nitrogen dioxide would be produced from 3.6 moles of oxygen
How many moles of oxygen would be needed to produce 4.67 mol nitrogen dioxide?
How many moles of nitrogen monoxide would be needed to produce 4.67 mol nitrogen dioxide?

9. Building off that…
If you can do mol to mol, the rest should be child’s play
Mass to mol
Convert mass to mol, then mol to mol.
The end
Mass to mass
Convert mass of given to mol, convert mols of given to moles of wanted, convert mol to mass
Particles to anything
Use Avagotto’s number (6.022 x 1023 particles/mol)

10. Mass to mol
In a reaction between the elements aluminum and chlorine, aluminum chloride is produced. How many grams of AlCl3 will be produced if 2.50 moles of Al react?
Al + Cl2  AlCl3 Unbalanced
2Al + 3Cl2  2AlCl3 Balanced

11. Mol to mass continued
2Al + 3Cl2  2AlCl3 How many grams of AlCl3 will be produced if 2.50 moles of Al react? Therefore g of AlCl3 will be produced by the reaction of 2.50 mol of Al
g AlCl3
2.50 mol AlCl3
1 mol AlCl3

12. Practice
The ammonia (NH3) used to make fertilizers for lawns and gardens is made by reacting nitrogen and hydrogen.
Determine the mass in grams of NH3 formed from 1.34 moles of nitrogen.
What is the mass in grams of hydrogen required to react with 1.34 moles of nitrogen?
How many moles of nitrogen are required to produce 11.7 grams of NH3?

13. Mass to mass
When nitrogen and hydrogen react, they form ammonia gas, which has the formula NH3. If 56.0 g of nitrogen are used up in the reaction, how many grams of ammonia will be produced?
N2 + H2  NH3 Unbalanced
N2 + 3H2  2NH3 Balanced

14. Therefore, 68.11 g of NH3 are produced from 56 g of N2
When nitrogen and hydrogen react, they form ammonia gas, which has the formula NH3. If 56.0 g of nitrogen are used up in the reaction, how many grams of ammonia will be produced?
N2 + 3H2  2NH3
Convert grams to mol
Convert mol to gram
Therefore, g of NH3 are produced from 56 g of N2
mol NH3
68.11 g NH3

15. Particles to Particles
Hydrogen and Oxygen react to produce water. How many particles of hydrogen would be needed to react fully with 1.01 x 1012 particles of oxygen?
H2+ O2  H2O Unbalanced
2H2 + O2  2H2O Balanced

16. Convert particles to mol Convert mol to mol Convert mol to particles
Hydrogen and Oxygen react to produce water. How many particles of hydrogen would be needed to react fully with 1.01 x 1012 particles of oxygen?
2H2 + O2  2H2O
Convert particles to mol
Convert mol to mol
Convert mol to particles
Therefore we expect to need 2.14 x 1012 particles of H2 to react fully with 1.01 x 1012 particles of O2
2.14 x 1012
particles of H2
x mol O2

17. Practice
Packet!

18. Percent yield
How much are you actually getting out of your experiment?
% Yield = Actual yield/theoretical yield x 100

19. Convert mass to mol Convert mol to mol Convert mol to mass 216g H2O
What is the total mass of H2O produced when 384g Cu is completely consumed in the following reaction? Cu + 4HNO3  Cu(NO3)2 + 2H2O + 2NO2
Convert mass to mol
Convert mol to mol
Convert mol to mass
Therefore we expect to produce 216g of H2O.
What is the percent yield if this procedure was carried out in the lab and only produced 200g of H2O
216g H2O

20. 92.6% = 200g 216g % yield = Actual yield Theoretical yield
We expected to produce 216g of H2O. What is the percent yield if this procedure was carried out in the lab and only produced 200g of H2O
What was the actual yield?
200g
What was the theoretical yield?
216g
Solve
Therefore, our % yield is 92.6%
200g
Actual yield
92.6% =
% yield =
Theoretical yield
216g

21. Practice
Ethanol (C2H5OH) is produced from the fermentation of sucrose in the presence of enzymes:
C12H22O11(aq) + H2O(g)  C2H4OH(l) + CO2(g)
determine the theoretical and percent yields of ethanol if 684g sucrose undergoes fermentation and 349g ethanol is obtained
T.Y. = 369g
% yield = 94.6%
Lead (II) oxide is obtained by roasting galena, lead (II) sulfide, in air:
PbS(s) + O2(g)  PbO(s) + SO2(g)
Determine the theoretical and percent yields of PbO if 200.0g PbS is heated and 170.0g PbO is obtained
T.Y. = 186.6g PbO
% Yield = 91.10%

22. Limiting reagent The reactant that is completely consumed during a rxn
Reactants that remain are excess reactants
Reagent

23. To find the limiting reagent
Compare the actual mol:mol ratio to the balanced formula’s mol:mol ratio
Calculations must be based on given amounts of limiting reactants

24. Fe + 2NiO(OH) + 2H2O  Fe(OH)2 + 2Ni(OH)2
Determine the number of moles of Fe(OH)2 produced if 5.00 mol Fe and 8.00 mol of NiO(OH) react. What is the limiting reactant? What reactant is in excess and by how many moles?
1mol Fe::2mol NiO(OH). Therefore,5mol Fe would need 10mol NiO(OH)
Since we only have 8.00mol NiO(OH), this is our limiting reagent.

25. Therefore 1 mol of Fe is in excess 4mol Fe
Fe + 2NiO(OH) + 2H2O  Fe(OH)2 + 2Ni(OH)2 Determine the number of moles of Fe(OH)2 produced if 5.00 mol Fe and 8.00 mol of NiO(OH) react. What is the limiting reactant? What reactant is in excess and by how many moles?
Since NiO(OH) is the limiting reagent, find how many moles of Fe are needed to completely react with the amount of it we do have
4 Mols of Fe are needed
Therefore 1 mol of Fe is in excess
4mol Fe
5.00mol Fe(provided) - 4mol Fe(needed) = 1mol Fe(excess)

26. Limiting Reagents It gets complicated
If you are given anything other then mol, you need to first convert to mol for BOTH reagents before you can determine the limiting factor!
Limiting
Reagents

27. Zn + 2MnO2 +H2O  Zn(OH)2 + Mn2O3
Determine the limiting reactant if 25.0g Zn and 30.0g MnO2 are used, Determine the mass of Zn(OH)2 produced.
0.77g moles of MnO2 required to react with 25.0g of Zn
0.17 moles of Zn required to react with 30.0g of Mn(OH)2
Mn(OH)2 is the limiting reagent
excess of 0.21mol Zn ( = 0.21)
Use 30.0g to find the mass of Zn(OH)2
17.07g Zn(OH)2