1. Chapter 4
Chemical Reactions

2. 4.1 Chemical Reactions and Chemical Equations

3. Chemical Equations
Concise representations of chemical reactions

4. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

5. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Reactants appear on the left side of the equation.

6. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Products appear on the right side of the equation.

7. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
The states of the reactants and products are written in parentheses to the right of each compound.

8. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation.

9. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule

10. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule
Coefficients tell the number of molecules

11. Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants and products

12. Hints for Balancing
If an element occurs in only one compound on each side of the equation, balance that first
Balance free elements last
Balance polyatomic ions as a group unit if they remain unchanged
If you get stuck, try doubling everything!

13. General Chemistry: Chapter 4
Balancing Equations
An equation can be balanced only by adjusting the coefficients of formulas.
Never introduce extraneous atoms to balance.
NO + O2 → NO2 + O
Never change a formula for the purpose of balancing an equation.
NO + O2 → NO3
Copyright © 2011 Pearson Canada Inc.
General Chemistry: Chapter 4

14. Example 4-1 A Balance the equations: H3PO4 + CaO  Ca3(PO4)2 + H2O
C3H8 + O2  CO2 + H2O

15. Example 4-2A
Write a balanced equation to represent the reaction of mercury(II) sulfide and calcium oxide to produce calcium sulfide, calcium sulfate and mercury metal.

16. (s) Solid (l) Liquid (g) Gas (aq) aqueous Over the arrow: catalyst, heat or conditions needed

17. 4.2 Chemical Equations and Stoichiometry
To Measure Elements!

18. Stoichiometric Calculations
From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

19. Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams

20. Switching Substances
YOU CAN ONLY SWITCH SUBSTANCES IN MOLES!! Must use balanced equation!

21. Example 4-3A
How many moles of O2 are produced from the decomposition of 1.76 moles of potassium chlorate?
2 KClO3(s)  2 KCl(s) + 3O2(g)

22. Example 4-4A
How many grams of magnesium nitride are produced by the reaction of 3.82 grams of Mg and excess N2?

23. Example 4-5A
For the reaction NH3 + O2  N2 + H2O How many grams of NH3 are consumed per gram of O2?

24. 4.3 Chemical Reactions in Solution

25. Molarity
Solvent: does the dissolving Solute: is dissolved Solution: Solute and Solvent Molarity = Amount of Solute (moles) Volume of Solution (Liters)

26. Example 4-8A
A 22.3 gram sample of acetone is dissolved in enough water to produced 1.25L of solution. What is the molarity of the acetone in this solution?

27. Example 4-9A
An aqueous solution saturated with NaNO3 at 25°C is 10.8 M NaNO3. What mass of NANO3 is present 125mL of this solution at 25°C?

28. Dilutions from Stock
M1V1 = M2V2

29. General Chemistry: Chapter 4
Solution Dilution
Mi × Vi = ni
Mi  Vi
Mf  Vf
= nf = Mf × Vf
M = n V
Mi × Vi
Mf = Vf
= Mi Vi Vf
Figure 4-6
Visualizing the dilution of a solution
Copyright © 2011 Pearson Canada Inc.
General Chemistry: Chapter 4

30. Example 4-10A
A 15.00mL sample of 0.450M K2CrO4 is diluted to mL. What is the concentration of the new solution?

31. Example 4-11A
How many milliliters of M K2CrO4 must be added to excess AgNO3(aq) to produce 1.50 g Ag2CrO4?

32. 4.4 Determining the Limiting Reactant
Limiting Reactants

33. How Many Cookies Can I Make?
You can make cookies until you run out of one of the ingredients
Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

34. How Many Cookies Can I Make?
In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

35. Limiting Reactants
The limiting reactant is the reactant present in the smallest stoichiometric amount

36. Limiting Reactants
The limiting reactant is the reactant present in the smallest stoichiometric amount
In other words, it’s the reactant you’ll run out of first (in this case, the H2)

37. Limiting Reactants
In the example below, the O2 would be the excess reagent

38. Example 4-12A
If 215 g P4 is allowed to react with 725 g Cl2 how many grams of PCl3 are formed?
P4 + 6Cl2  4PCl3

39. Example 13A
From 12A, which reactant is in excess and what mass of the reactant remains after the reaction to produce PCl3?

40. 4.5 Other Practical Matters in Reaction Stoichiometry

41. Theoretical Yield
The theoretical yield is the amount of product that can be made
In other words it’s the amount of product possible as calculated through the stoichiometry problem
This is different from the actual yield, the amount one actually produces and measures

42. Percent Yield Actual Yield Theoretical Yield Percent Yield = x 100
A comparison of the amount actually obtained to the amount it was possible to make
Actual Yield
Theoretical Yield
Percent Yield = x 100

43. Example I2 is limiting and 2.57 g is the theoretical yield
1.20 g Al reacts with 2.40 g I2. What is the limiting reactant?
2Al + 3 I2  2AlI3
1st find the reactant that produces the smallest amount of product.
1.20 g Al x 1mol Al x 2 mol AlI3 x g AlI3 = 18.1 g AlI3
27 g Al mol Al mol AlI3
2.40 g I2 x 1 Mol I2 x 2 mol AlI3 x g AlI3 = 2.57 g AlI3
253.8 g I mol I mol AlI3
I2 is limiting and 2.57 g is the theoretical yield

44. Theoretical Yield % Yield= actual Yield/ theoretical yield x 100
Say the actual yield was 2.05 g
Then