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Chapter 4 Chemical Reactions. 4.1 Chemical Reactions and Chemical Equations.

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Presentation on theme: "Chapter 4 Chemical Reactions. 4.1 Chemical Reactions and Chemical Equations."— Presentation transcript:

1 Chapter 4 Chemical Reactions

2 4.1 Chemical Reactions and Chemical Equations

3 3 Chemical Equations Concise representations of chemical reactions

4 4 Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

5 5 Anatomy of a Chemical Equation Reactants appear on the left side of the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

6 6 Anatomy of a Chemical Equation Products appear on the right side of the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

7 7 Anatomy of a Chemical Equation The states of the reactants and products are written in parentheses to the right of each compound. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

8 8 Anatomy of a Chemical Equation Coefficients are inserted to balance the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

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

10 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 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 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4Slide 13 of 24 Balancing Equations Never introduce extraneous atoms to balance. NO + O 2 → NO 2 + O Never change a formula for the purpose of balancing an equation. NO + O 2 → NO 3 An equation can be balanced only by adjusting the coefficients of formulas.

14 Example 4-1 A Balance the equations: H 3 PO 4 + CaO  Ca 3 (PO 4 ) 2 + H 2 O C 3 H 8 + O 2  CO 2 + H 2 O

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 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 19 Stoichiometric Calculations Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 … use the coefficients to find the moles of H 2 O… and then turn the moles of water to grams C 6 H 12 O O 2  6 CO H 2 O

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

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

22 Example 4-4A How many grams of magnesium nitride are produced by the reaction of 3.82 grams of Mg and excess N 2 ?

23 Example 4-5A For the reaction NH 3 + O 2  N 2 + H 2 O How many grams of NH 3 are consumed per gram of O 2 ?

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 NaNO 3 at 25°C is 10.8 M NaNO 3. What mass of NANO 3 is present 125mL of this solution at 25°C?

28 Dilutions from Stock M 1 V 1 = M 2 V 2

29 Solution Dilution Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4Slide 29 of 24 Figure 4-6 Visualizing the dilution of a solution M i × V i = n i M i  V i M f  V f = n f = M f × V f M = n V M i × V i M f = VfVf = M i ViVi VfVf

30 Example 4-10A A 15.00mL sample of 0.450M K 2 CrO 4 is diluted to mL. What is the concentration of the new solution?

31 Example 4-11A How many milliliters of M K 2 CrO 4 must be added to excess AgNO 3(aq) to produce 1.50 g Ag 2 CrO 4 ?

32 4.4 Determining the Limiting Reactant Limiting Reactants

33 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 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 35 Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount

36 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 H 2 )

37 37 Limiting Reactants In the example below, the O 2 would be the excess reagent

38 Example 4-12A If 215 g P 4 is allowed to react with 725 g Cl 2 how many grams of PCl 3 are formed? P 4 + 6Cl 2  4PCl 3

39 Example 13A From 12A, which reactant is in excess and what mass of the reactant remains after the reaction to produce PCl 3 ?

40 4.5 Other Practical Matters in Reaction Stoichiometry

41 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 42 Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual Yield Theoretical Yield Percent Yield =x 100

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

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


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