1. USING THE REACTION EQUATION LIKE A RECIPE
STOICHIOMETRY
USING THE REACTION EQUATION LIKE A RECIPE

2. Nearly everything we use is manufactured from chemicals.
USING EQUATIONS
Nearly everything we use is manufactured from chemicals.
Soaps, shampoos, conditioners, cd’s, cosmetics, medications, and clothes.
For a manufacturer to make a profit the cost of making any of these items can’t be more than the money paid for them.
Chemical processes carried out in industry must be economical, this is where balanced equations help.

3. Equations are a chemist’s recipe.
USING EQUATIONS
Equations are a chemist’s recipe.
Eqs tell chemists what amounts of reactants to mix and what amounts of products to expect.
When you know the quantity of one substance in a rxn, you can calculate the quantity of any other substance consumed or created in the rxn.
Quantity meaning the amount of a substance in grams, liters, molecules, or moles.

4. USING EQUATIONS
The calculation of quantities in chemical reactions is called stoichiometry.
Imagine you are in charge of manu-facturing for Rugged Rider Bicycle Company.
The business plan for Rugged Rider requires the production of 128 custom-made bikes each day.
You are responsible for insuring that there are enough parts at the start of each day.

5. F +S+2W+H+2P  FSW2HP2 USING EQUATIONS
Assume that the major components of the bike are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P).
The finished bike has a “formula” of FSW2HP2.
The balanced equation for the production of 1 bike is.
F +S+2W+H+2P 
FSW2HP2

6. # of wheels = ? wheels USING EQUATIONS
Now in a 5 day workweek, Rugged Riders is scheduled to make 640 bikes. How many wheels should be in the plant on Monday morning to make these bikes?
What do we know?
Number of bikes = 640 bikes
1 FSW2HP2=2W (balanced eqn)
What is unknown?
# of wheels = ? wheels

7. The connection between wheels and bikes is 2 wheels per bike
The connection between wheels and bikes is 2 wheels per bike. We can use this information as a conversion factor to do the calculation.
2 W
640 FSW2HP2 =
Wheels
1 FSW2HP2
We can make the same kinds of connections from a chemical rxn eqn.
N2(g) + 3H2(g)  2NH3(g)
The key is the “coefficient ratio”.

8. 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chem rxn.
1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
N2 and H2 will always react to form ammonia in this 1:3:2 ratio of moles.
So if you started with 10 moles of N2 it would take 30 moles of H2 and would produce 20 moles of NH3

9. Using the coefficients, from the balanced rxn equation as ratios to make connections between reactants and products, is the most important information that a rxn equation provides.
Using this information, you can calculate the amounts of the reactants involved and the amount of product you might expect.
Any calculation done with the next process is a theoretical value, the real world isn’t always perfect.

10. MOLE – MOLE EXAMPLE
The following rxn shows the synthesis of aluminum oxide.
3O2(g) + 4Al(s)  2Al2O3(s)
3O2(g) + 4Al(s)  2Al2O3(s)
If you only had 1.8 mols of Al how much product could you make?
Given: 1.8 moles of Al
Uknown: ____ moles of Al2O3

11. MOLE – MOLE EXAMPLE
Solve for the unknown:
3O2(g) + 4Al(s)  2Al2O3(s)
2 mol Al2O3
1.8 mol Al
= 0.90mol
Al2O3
4 mol Al
Mole Ratio

12. MOLE – MOLE EXAMPLE 2
The following rxn shows the synthesis of aluminum oxide.
3O2(g) + 4Al(s)  2Al2O3(s)
If you wanted to produce 24 mols of product how many mols of each reactant would you need?
Given: 24 moles of Al2O3
Uknown: ____ moles of Al
____ moles of O2

13. MOLE – MOLE EXAMPLE 2
Solve for the unknowns:
3O2(g) + 4Al(s)  2Al2O3(s)
4 mol Al
24 mol Al2O3
= 48 mol Al
2 mol Al2O3
3 mol O2
24 mol Al2O3
= 36 mol O2
2 mol Al2O3

14. Practice for You…
How many mols of hydrogen will be produced if 0.44 mol of CaH2 reacts according to the following equation? CaH2 + 2H2O  Ca(OH) H2
Iron will react with oxygen to produce Iron III oxide. How many moles of Iron III oxide will be produced if 0.18 mol of Iron reacts?
(.89 mol H2)
(.090 mol Fe2O3)

15. MASS – MASS CALCULAT’NS
No lab balance measures moles directly, generally mass is the unit of choice.
From the mass of 1 reactant or prod-uct, the mass of any other reactant or product in a given chemical equation can be calculated, provided you have a balanced rxn equation.
As in mole-mole calcs, the unknown can be either a reactant or a product.

16. MASS – MASS CALCULAT’NS 1
Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2).
CaC2 + 2H2O  C2H2 + Ca(OH)2
CaC2 + 2H2O  C2H2 + Ca(OH)2
How many grams of C2H2 are produced by adding water to 5.00 g CaC2?

17. MASS – MASS CALCULAT’NS 1
What do we know?
Given mass = 5.0 g CaC2
Mole ratio: 1 mol CaC2 = 1 mol C2H2
MM of CaC2 = 64.0 g CaC2
MM of C2H2 = 26.0g C2H2
What are we asked for?
grams of C2H2 produced

18. 1 mol CaC2 5.0 g CaC2 = .07813 mol CaC2 64.0 g CaC2 1 mol C2H2
Step 1: “Get to Moles!” in this case that can be done by using the Molar Mass of your given compound.
1 mol CaC2
5.0 g CaC2
= mol
CaC2
64.0 g CaC2
Step 2: Now we are ready for the KEY step…converting from mols of our given to mols of unknown using the mole ratio.
1 mol C2H2
mol
CaC2
= mol
C2H2
1 mol CaC2

19. Step 3: Since we are asked for mass of our unknown in this problem, we need to use our molar mass of our unknown and convert our newly calculated mols into grams.
26.0 g C2H2
mol
C2H2
= 2.03 g C2H2
1 mol C2H2
Summary of 3 Steps:
Get to Moles
Mole Ratio
Get to desired final unit

20. MASS – MASS CALCULAT’NS 2
The double replacement reaction between Lead II nitrate and Potassium Iodide produces a bright yellow precipitate that can be used as a color additive in paint. How many grams of Potassium iodide would we need to completely react 25.3 g of Lead II nitrate?
Pb(NO3)2 + 2 KI  PbI KNO3
Pb(NO3)2 + 2 KI  PbI KNO3
mass A  mols A  mols B  mass B

21. MASS – MASS CALCULAT’NS 2
331.2g Pb(NO3)2
1mol Pb(NO3)2
25.3 g Pb(NO3)2
1mol Pb(NO3)2
2mol KI
1mol KI
166 g KI
= 25.4 g KI

22. Practice for You…
What mass of Barium chloride is needed to react completely with g of Sodium phosphate accor-ding to the following rxn equation? BaCl2 + Na3PO4  Ba3(PO4)2 + NaCl
Use the equation to determine what mass of FeS must react to form g of FeCl FeS + HCl  H2S + FeCl2
(89.2 g BaCl2)
(226 g FeS)

23. A balanced reaction equation indicates the relative numbers of moles of reactants and products.
We can expand our stoichiometric calculations to include any unit of measure that is related to the mole.
The given quantity can be expressed in numbers of particles, units of mass, or volumes of gases at STP.
The problems can include mass-volume, volume-volume, and particle-mass calculations.

24. In any of these problems, the given quantity is first converted to moles.
Then the mole ratio from the balanced eqn is used to convert from the moles of given to the number of moles of the unknown
Then the moles of the unknown are converted to the units that the problem requests.
The next slide summarizes these steps for all typical stoichiometric problems

26. 2H2O  2H2 + O2 MORE MOLE EXAMPLES
How many molecules of O2 are produced when a sample of 29.2 g of H2O is decomposed by electrolysis according to this balanced equation:
2H2O  2H2 + O2

27. What are we asked for? MORE MOLE EXAMPLES What do we know?
Mass of H2O = 29.2 g H2O
2 mol H2O = 1 mol O2 (from balanced equation)
MM of H2O = 18.0 g H2O
1 mol O2 = 6.02x1023 molecules of O2
What are we asked for?
molecules of O2

28. mass A  mols A  mols B 
molecules B
29.2 g H2O
1 mol H2O
1 mol O2
18.0 g H2O
2 mol H2O
6.02x1023 molecules O2
1 mol O2
= 4.88 x 1023
molecules O2

29. MORE MOLE EXAMPLES
The last step in the production of nitric acid is the reaction of NO2 with H2O.
3NO2+H2O2HNO3+NO
How many liters of NO2 must react with water to produce 5.00x1022 molecules of NO?

30. MORE MOLE EXAMPLES What do we know? What are we asked for?
Molecules NO = 5.0x1022 molecules NO
1 mol NO = 3 mol NO2 (from balanced equation)
1 mol NO = 6.02x1023 molecules NO
1 mol NO2 = 22.4 L NO2
What are we asked for?
Liters of NO2

31. molecules A mols mols B volume B
1 mol NO
5.0x1022 mol-ecules NO
3 mol NO2
1 mol NO
6.02x1023 mol-ecules NO
22.4 L NO2
1 mol NO2
= 5.58 L NO2

32. Aspirin can be made from a chemical rxn between the reactants salicylic acid and acetic anhydride. The products of the rxn are acetyl-salicylic acid (aspirin) and acetic acid (vinegar). Our factory makes 125, count bottles of Bayer Aspirin/day. Each bottle contains 100 tablets, and each tablet contains 325mg of aspirin. How much in kgs + 10% for production problems, of each reactant must we have in order to meet production?
C7H6O3 + C4H6O3  C9H8O4 + HC2H3O2
Salicylic acid
Acetic anhydride
aspirin
vinegar

33. Mass of salicylic acid in kgs + 10%
What do we know?
Make 125,000 aspirin bottles/day
100 aspirin/bottle
325 mg aspirin/tablet
Mole ratio of aspirin to salicylic acid (1:1) and acetic anhydride (1:1)
MM aspirin = g
MM C7H6O3 = g
MM C4H6O3 = g
What are we asked for?
Mass of salicylic acid in kgs + 10%
Mass of acetic anhydride in kgs + 10%

34. 100 tablets 1 bottle 1 tablet 1000 mg 180.16g = 22,549.4 mols aspirin
325mg asp.
125,000 bottles
1 bottle
1 tablet
1 g
1mol asp.
1000 mg
180.16g
= 22,549.4 mols aspirin

35. = 3068.97 kg salicylic acid + (306.897 g) 1000 g
1 mol C7H6O3
136.10g C7H6O3
22,549.4 mols aspirin
1 mol asp
1 mol C7H6O3
1 kg
= kg salicylic acid + ( g)
1000 g
= 3380 kg of salicylic acid

36. Acetic Anhydride: 1 mol asp 1 mol C4H6O3 = 2301.39 kg
102.06g C4H6O3
22,549.4 mols aspirin
1 mol asp
1 mol C4H6O3
1 kg
= kg
Acetic anhydride kg
1000 g
= 2530 kg Acetic anhydride