1. Balancing Equations and Stoichiometry
Chapter 4:
Balancing Equations and Stoichiometry

2. Key Terms & Concepts Stoichiometry Chemical Equations
reactants and products
balancing chemical equations
Chemical Calculations
Limiting Reactant
Theoretical and Percent Yield

3. Stoichiometry
Stoichiometry is the study of the quantitative nature of chemical formulas and chemical reactions.
Stoichiometry is one the the most essential tools in chemistry
It allows to quantify everything from global warming to drug manufacturing

4. Chemical Equations
Chemical reactions are represented in a concise manner by chemical equation
For example, when H2 burns in O2, H2O is formed.
The chemical equation for this reaction is:
2 H2 + O2  2H2O

5. Chemical Equations 2 H2 + O2  2 H2O
The compounds on the left of the arrow are called “reactants”
The compounds on the right of the arrow are called “products”
H2 and O2 are reactants, H2O is the product

6. Chemical Equations All chemical equations must meet this requirement
2 H2 + O2  2 H2O
Notice that the number of atoms or each element is equal on both sides of the equation
4 H, 2 0
All chemical equations must meet this requirement
Chemical equations must be balanced!!
We balance equations by changing coefficients, not chemical formulas

7. Chemical Equations

8. Chemical Equations Consider the following chemical equation
CH4 + O2  CO2 + H2O
unbalanced
Start with elements that only appear in one compound on either side of the equation
C and H are only in one compound on either side
C is balanced

9. Chemical Equations CH4 + O2  CO2 + H2O unbalanced
4 H’s in reactants, 2 H’s in products
Put coefficient of 2 in front of H2O
4 H’s in reactants, 4 H’s in products
CH4 + O2  CO H2O

10. Chemical Equations CH4 + O2  CO2 + 2 H2O unbalanced
2 O’s in reactants, 4 O’s in products
Put coefficient of 2 in front of O2
4 O’s in reactants, 4 O’s in products
CH O2  CO H2O
balanced

11. Chemical Equations

12. Chemical Equations Consider this equation: C3H8 + O2  CO2 + H2O
3C + 8H + 2O  1C + 2H + 3O

13. Chemical Equations Balance C and H C3H8 + O2  3CO2 + 4H2O
3C + 8H + 2O  3C + 8H + 10O

14. Chemical Equations Balance O C3H8 + 5O2  3CO2 + 4H2O
3C + 8H + 10O  3C + 8H + 10O

15. Chemical Equations (5) C8H18 + O2  CO2 + H2O Example 4.1
Balance the following chemical equations
(1) Mg + HCl  MgCl2 + H2
(2) K + H2O  KOH + H2
(3) CaCl2 + Na3PO4  Ca3(PO4)2 + NaCl
(4) NaN3  Na + N2
(5) C8H O2  CO2 + H2O

16. Chemical Equations (5) 2 C8H18 + 25 O2  16 CO2 + 18 H2O Example 4.1
Balance the following chemical equations
(1) Mg HCl  MgCl2 + H2
(2) 2 K H2O  2 KOH + H2
(3) 3 CaCl Na3PO4  Ca3(PO4) NaCl
(4) 2 NaN3  2 Na N2
(5) 2 C8H O2  CO H2O

17. Chemical Equations Example 4.2
Write a balanced chemical equation for the following reactions
ammonium nitrate decomposes to nitrogen gas, oxygen gas, and water
iron reacts with oxygen gas and water to form iron(II) hydroxide
ammonia reacts with oxygen gas to produce nitrogen monoxide and water

18. Chemical Equations Example 4.2
Write a balanced chemical equation for the following reactions
2 NH4NO3  2 N2 + O2 + 4 H2O
2 Fe + O2 + 2 H2O  2 Fe(OH)2
4NH3 + 5 O2  4 NO + 6 H2O

19. Chemical Calculations
2 H O2  2 H2O
2 molecules 1 molecule 2 molecules
2(6.022x1023) molecules x1023 molecules (6.022x1023) molecules
2 mol mol 2 mol
Stoichiometric coefficients can be interpreted as either number of molecules or number of moles.

20. Chemical Calculations
Example 4.3
How many moles of water can be produced from 5.25 mol O2?

21. Chemical Calculations
Example 4.4
How many moles of oxygen are required to completely react with 8.50 moles of butane, C4H10?

22. Chemical Calculations
We can’t directly measure moles. We can measure mass.
We can use the stoichiometric coefficients of a reaction to determine the mass relationships.
However, we must always convert mass to moles.
We cannot directly compare the masses of reactants and products.
We can only compare the moles of reactants and products.

23. Chemical Calculations
The general scheme is:

24. Chemical Calculations
Example 4.5
Geranyl formate is used as a synthetic rose essence in cosmetics. The compound is prepared from formic acid and geraniol:
HCOOH + C10H18O  C11H18O2 + H2O
A chemist needs to make some geranyl formate for a batch of perfume. How many grams of geranyl formate can a chemist make from 375g of geraniol?

25. Chemical Calculations

26. Chemical Calculations
Example 4.6
Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide. The lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by each 1.00 g of lithium hydroxide?
2 LiOH (s) + CO2 (g)  Li2CO3 (g) + H2O (l)

27. Chemical Calculations
Example 4.6
2 LiOH (s) + CO2 (g)  Li2CO3 (g) + H2O (l)

28. Theoretical and Percent Yield
The amount of product that can be produced from a given amount of reactants is the theoretical yield.
However, no reaction goes to actual completion. The amount of products that is actually produced from a given amount of reactants is the actual yield.
Some reactants may not react
Reactants may react in an undesired way (side reactions)
May be difficult to remove products from pot

29. Theoretical and Percent Yield
The extent of the desired reaction is typically reported as the percent yield.

30. Theoretical and Percent Yield
Example 4.7
Look back at Example If the chemist starts with 375g of geraniol and collects 417g of purified product, what is the percent yield of the synthesis?

31. Theoretical and Percent Yield
Example 4.8
25.0 g of sodium metal is burned in an excess of chlorine gas. What is the theoretical yield of sodium chloride? If 54.8 g of sodium chloride is actually produced, what is the percent yield of the reaction?
2 Na + Cl2  2 NaCl

32. Theoretical and Percent Yield

33. Theoretical and Percent Yield

34. Theoretical and Percent Yield
Example 4.9
Titanium is a strong, lightweight, corrosion-resistant metal that is used in aeronautics and bicycle frames. It is prepared by the reaction of titanium (IV) chloride with molten magnesium between 950C and 1150C.
TiCl4(g) + 2 Mg(l)  Ti(s) + 2 MgCl2(l)
In a certain industrial process 3.54x107 g of TiCl4 are reacted with 1.13x107 g of Mg. (a)Calculate the theoretical yield of Ti in grams. (b)Calculate the percent yield if 7.91x106 g of Ti are actually produced.

35. Theoretical and Percent Yield

36. Theoretical and Percent Yield

37. Limiting Reactants
Most reactions do not occur with stoichiometric equivalent amounts of each reactant.
One reactant is used up first
This reactant is the limiting reactant because it limits the amount of products that can be formed

38. Limiting Reactants Consider the “ham sandwich” example
one sandwich is made from one slice of ham, one slice of cheese and two slices of bread
How many ham sandwiches can be made from six slices of ham, seven slices of cheese and 14 slices of bread?
What is the limiting reactant?

39. Limiting Reactants

40. Limiting Reactants
If a problem gives specific amounts of two or more reactants it is a limiting reactant problem.
Determine the amount of product that can be formed from each reactant
The reactant which produces the smallest amount of product is the limiting reactant
The remaining reactants are said to be in excess

41. Limiting Reactants Example 4.10
How many moles of water can be formed when 10.0 moles of H2 reacts with 4.50 moles of O2? What is the limiting reactant?

42. Limiting Reactants
Example 4.10
2 H2 + O2  2 H2O

43. Limiting Reactants Example 4.11
Solutions of sulfuric acid and lead (II) acetate react to form solid lead (II) sulfate and aqueous acetic acid. If 15.0 g of sulfuric acid and 15.0 g of lead (II) acetate are mixed, calculate the number of grams of lead (II) sulfate that can be produced. Also calculate the number of grams of the excess reagent remaining after the reaction is completed.

44. H2SO4 + Pb(CH3COO)2  PbSO4 + 2 CH3COOH
Limiting Reactants
H2SO4 + Pb(CH3COO)2  PbSO4 + 2 CH3COOH

45. Limiting Reactants

46. End-of-Chapter Exercises
Suggested End-of-Chapter Exercises
5, 9, 10, 14, 17, 20, 22, 28