1. Stoichiometry Chemical Equations
Dr. Ron Rusay
Spring 2004
© Copyright 2004 R.J. Rusay

2. Chemical Reactions Atoms, Mass & Balance: eg. Fe + S -->

3. Chemical Equation Representation of a chemical reaction:
_ C2H5OH + _ O2  _ CO2 + _ H2O
reactants products
C=2; H =5+1=6; O=2+1 C=1; H=2; O=2+1
1 C2H5OH O2  2 CO H2O

4. Chemical Equation C2H5OH + 3 O2  2 CO2 + 3 H2O
The equation is balanced and the reaction can be completely stated as:
1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.

5. The Chemical Equation: Mole & Masses
C2H5OH O2  2 CO H2O
46g (1 mole) of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.
How many grams of carbon dioxide and water are respectively produced?

6. The Chemical Equation: Moles & Masses (Alcohol)
C2H5OH O2  2 CO H2O
46g (1 mole) of ethanol requires 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.
How many grams of oxygen are needed to react with all of the 46g (1 mole) of ethanol ?
How many grams of oxygen are needed to react with 15.3g of ethanol in a 12oz. beer ?

9. Gases WKS [Extra Credit] An average pair of human lungs contains about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm, determine a) the number of moles of O2 and b) the mass of O2 and the total mass of air in a typical breath.
How many breaths would it take to oxidize the alcohol in 1 beer if the oxygen inhaled was used only for that purpose?

10. Limiting Reagent An Ice Cream Sundae

11. Limiting Reagent

12. Mass Applications: Limiting Reagent
How many FeS molecules are produced in each case?
What is left over?
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13. Mass Applications: Limiting Reagent
How many ZnS molecules can be produced in each case?
What is left over?
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14. Mass Applications: Limiting Reagent
How do masses of reactants relate? Is there enough mass of each reactant for the reaction to consume all of both of them or will there be some left of one of them?
C2H5OH O2  2 CO H2O
What would happen if 24. g of O2 (0.75 mol) were available for the reaction of 15.3 g of ethanol (0.33 mol)?
How much O2 is needed to react with 0.33 mol of ethanol?
= 0.33 mol C2H5OH x 3 mol O2 / 1 mol C2H5OH
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15. Mass Applications: Determining a Limiting Reagent
One of the reactants, oxygen, has fewer stoichiometrically adjusted moles than the other reactant, ethanol.
Therefore, the reactant with the smaller value is the limiting reagent. In this case oxygen.
Not all of the ethanol can react. Once the oxygen is depleted, the chemical reaction stops until more oxygen becomes available.
The unreacted amount of alcohol remains in the blood.

21. Mass Calculations: Products Reactants
1. Balance the chemical equation.
2. Convert mass of reactant(s) or product(s)
to moles.
3. Identify mole ratios in balanced equation:
They serve as the “Gatekeeper”.
4. Calculate moles of desired product or
reactant. (Use the Limiting Reagent as basis)
5. Convert moles to grams.
© Copyright R.J. Rusay

22. Mass Calculations: Products Reactants
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23. Mass Calculations: Products Reactants
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24. Mass Calculations: Reactants Products
How many grams of salicylic acid are needed to produce 1.80 kg of aspirin?
Balanced Equation:

25. Mass Calculations: Reactants Products
grams (Aspirin)
grams (Salicylic Acid)
Avogadro's Number
Atoms
Molecules
Stoichiometry
grams (SA)
(Molecular
1800 grams (A)
1 mol (A)
1 mol SA
Weight SA)
1380 grams (SA)
1 mol A
grams (A)
1 mol (SA)
(Molecular
Weight A)
"Gatekeeper"
Calculation:
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26. Percent Yield
In synthesis, the actual yield (g) is measured and compared to the theoretical yield (g). This is the percent yield: % = actual / theoretical x 100
If a reaction produced 2.45g of Ibogaine, C20H26N2O, a natural product with strong promise in treating heroin addiction, and the theoretical yield was 3.05g, what is the % yield?
© Copyright R.J. Rusay

27. Percent Yield
A reaction was conducted that theoretically would produce moles of quinine, C20 H24 N2 O2 . The actual amount of isolated quinine was 780 mg. What is the percent yield of quinine?
324 g/mol x mol = 81g = 810mg(theoretical)
% Yield = 780 mg/ 810 mg x 100
% Yield = 96%