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Dr. S. M. Condren Chapter 4 Chemical Equations and Stoichiometry.

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Presentation on theme: "Dr. S. M. Condren Chapter 4 Chemical Equations and Stoichiometry."— Presentation transcript:

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2 Dr. S. M. Condren Chapter 4 Chemical Equations and Stoichiometry

3 Dr. S. M. Condren CHEMICAL REACTIONS Reactants: Zn + I 2 Product: Zn I 2

4 Dr. S. M. Condren Chemical Equations Depict the kind of reactants and products and their relative amounts in a reaction. 4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s) The numbers in the front are called stoichiometric coefficients The letters (s), (g), and (l) are the physical states of compounds, (aq) refers to an aqueous or water solution.

5 Dr. S. M. Condren The Mole and Chemical Reactions: The Nano-Macro Connection 2 H 2 + O > 2 H 2 O 2 H 2 molecules 1 O 2 molecule 2 H 2 O molecules 2 H 2 moles molecules 1 O 2 mole molecules 2 H 2 O moles molecules 4 g H 2 32 g O 2 36 g H 2 O

6 Dr. S. M. Condren Stoichiometry stoi·chi·om·e·try noun 1. Calculation of the quantities of reactants and products in a chemical reaction. 2. The quantitative relationship between reactants and products in a chemical reaction.

7 Dr. S. M. Condren 4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s) This equation means 4 Al atoms + 3 O 2 molecules ---give---> 2 molecules of Al 2 O 3 4 moles of Al + 3 moles of O 2 ---give---> 2 moles of Al 2 O 3 Chemical Equations

8 Dr. S. M. Condren Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change. The Law of the Conservation of Matter Demo of conservation of matter 2HgO(s) ---> 2 Hg(liq) + O 2 (g) Chemical Equations

9 Dr. S. M. Condren Combination Reaction PbNO 3(aq) + K 2 CrO 4(aq) PbCrO 4(s) + 2 KNO 3(aq) Colorless yellow yellow colorless

10 Dr. S. M. Condren Because of the principle of the conservation of matter, an equation must be balanced. It must have the same number of atoms of the same kind on both sides. Chemical Equations Lavoisier, 1788

11 Dr. S. M. Condren ___ Al(s) + ___ Br 2 (liq) ---> ___ Al 2 Br 6 (s) Balancing Equations

12 Dr. S. M. Condren ____C 3 H 8 (g) + _____ O 2 (g) ----> _____CO 2 (g) + _____ H 2 O(g) ____B 4 H 10 (g) + _____ O 2 (g) ----> ___ B 2 O 3 (g) + _____ H 2 O(g) Balancing Equations

13 Dr. S. M. Condren EXAMPLE How much H 2 O, in moles results from burning an excess of H 2 in 3.3 moles of O 2 ? H 2 + O > H 2 O #mol H 2 O = 22 (3.3 mol O 2 )(2 mol H 2 O) (1 mol O 2 ) Now we need the stoichiometric factor = 6.6 mol H 2 O

14 Dr. S. M. Condren Stoichiometric Roadmap

15 Dr. S. M. Condren EXAMPLE How much H 2 O, in grams results from burning an excess of H 2 in 3.3 moles of O 2 ? H 2 + O > H 2 O #g H 2 O = 22 (3.3 mol O 2 )(2 mol H 2 O) (1 mol O 2 ) = 1.2x10 2 g H 2 O (18.0 g H 2 O) (1 mol H 2 O)

16 Dr. S. M. Condren EXAMPLE How much H 2 O, in grams results from burning an excess of H 2 in 3.3 grams of O 2 ? H 2 + O > H 2 O #g H 2 O = 22 (1 mol O 2 )(2 mol H 2 O) (1 mol O 2 ) = 3.7 g H 2 O (18.0 g H 2 O) (1 mol H 2 O) (3.3 g O 2 ) (32.g O 2 )

17 Dr. S. M. Condren Thermite Reaction

18 Dr. S. M. Condren Thermite Reaction Fe 2 O 3(s) + 2Al (s) Al 2 O 3(s) + 2 Fe (l)

19 Dr. S. M. Condren Thermite Reaction

20 Dr. S. M. Condren EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together?

21 Dr. S. M. Condren EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron. The mass of iron in 1 inch of this rail is: #g/in = (132 #/yard) (1 yard/36 in) (454 g/#) = 1.67 X 10 3 g/in The mass of iron in a weld adding 10% mass: #g = (1.67 X 10 3 g) (0.10) = 167 g

22 Dr. S. M. Condren EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? The mass of iron in a weld adding 10% mass: #g = (1.67 X 103 g) (0.10) = 167 g Balanced chemical equation: Fe 2 O Al ---> 2 Fe + Al 2 O 3 What mass of Fe 2 O 3 is required for the thermite process? #g Fe 2 O 3 = (167 g Fe) (1 mol Fe) (55.85 g Fe) (1 mol Fe 2 O 3 ) (2 mol Fe) (159.7 g Fe 2 O 3 ) (1 mol Fe 2 O 3 ) = 238 g Fe 2 O 3

23 Dr. S. M. Condren EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? The mass of iron in a weld adding 10% mass: #g = (1.67 X 103 g) (0.10) = 167 g Balanced chemical equation: Fe 2 O Al ---> 2 Fe + Al 2 O 3 What mass of Al is required for the thermite process? #g Al = (167 g Fe) (1 mol Fe) (55.85 g Fe) (2 mol Al) (2 mol Fe) ( g Al) (1 mol Al) = 80.6 g Al

24 Dr. S. M. Condren EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe #g Fe 2 O 3 = 238 g Fe 2 O 3 #g Al = 80.6 g Al

25 Dr. S. M. Condren In a given reaction, there is not enough of one reagent to use up the other reagent completely.In a given reaction, there is not enough of one reagent to use up the other reagent completely. The reagent in short supply LIMITS the quantity of product that can be formed.The reagent in short supply LIMITS the quantity of product that can be formed. Reactions Involving a LIMITING REACTANT

26 Dr. S. M. Condren LIMITING REACTANTS eactants R eactantsProducts 2 NO(g) + O 2 (g) 2 NO 2 (g) Limiting reactant = ___________ Excess reactant = ____________

27 Dr. S. M. Condren EXAMPLE What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) balanced equation relates: 2Fe 2 S 3(S) 6H 2 O (l) 3O 2(g) have only: 1Fe 2 S 3 (S) 2H 2 O (l) 3O 2(g) not enough H 2 O to use all Fe 2 S 3 plenty of O 2

28 Dr. S. M. Condren EXAMPLE What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) if use all Fe 2 S 3 : (1.0 mol Fe 2 S 3 ) (4 mol Fe(OH) 3 ) #mol Fe(OH) 3 = (2 mol Fe 2 S 3 ) = 2.0 mol Fe(OH) 3

29 Dr. S. M. Condren EXAMPLE: What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) if use all H 2 O: (2.0 mol H 2 O) (4 mol Fe(OH) 3 ) #mol Fe(OH) 3 = (6 mol H 2 O) = 1.3 mol Fe(OH) 3

30 Dr. S. M. Condren EXAMPLE: What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) if use all O 2 (3.0 mol O 2 ) (4 mol Fe(OH) 3 ) #mol Fe(OH) 3 = (3 mol O 2 ) = 4.0 mol Fe(OH) 3

31 Dr. S. M. Condren EXAMPLE: What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) 1.0 mol Fe 2 S 3 => 2.0 mol Fe(OH) mol H 2 O => 1.3 mol Fe(OH) mol O 2 => 4.0 mol Fe(OH) 3 least amount Since 2.0 mol H 2 O will produce only 1.3 mol Fe(OH) 3, then H 2 O is the limiting reactant. Thus the maximum number of moles of Fe(OH) 3 that can be produced by this reaction is 1.3 moles.

32 Dr. S. M. Condren Theoretical Yield the amount of product produced by a reaction based on the amount of the limiting reactant

33 Dr. S. M. Condren Actual Yield amount of product actually produced in a reaction

34 Dr. S. M. Condren Percent Yield actual yield % yield = * 100 theoretical yield

35 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process

36 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00 kg Cl 2 ) #kg N 2 H 4 =

37 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00 kg Cl 2 ) (1000 g Cl 2 ) #kg N 2 H 4 = (1 kg Cl 2 ) metric conversion

38 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00) (1000 g Cl 2 ) (1 mol Cl 2 ) #kg N 2 H 4 = (1) (70.9 g Cl 2 ) molar mass

39 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1 mol Cl 2 ) #kg N 2 H 4 = (1)(70.9)

40 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1 mol Cl 2 )(1 mol N 2 H 4 ) #kg N 2 H 4 = (1) (70.9) (1 mol Cl 2 )

41 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1 mol N 2 H 4 ) #kg N 2 H 4 = (1) (70.9)(1)

42 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1 mol N 2 H 4 ) (32.0 g N 2 H 4 ) #kg N 2 H 4 = (1)(70.9) (1) (1 mol N 2 H 4 ) molar mass

43 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1) (32.0 g N 2 H 4 )(1 kg N 2 H 4 ) #kg N 2 H 4 = (1)(70.9)(1)(1) (1000 g N 2 H 4 ) metric conversion

44 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1) (32.0)(1 kg N 2 H 4 ) #kg N 2 H 4 = (1)(70.9)(1)(1)(1000)

45 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1) (32.0)(1 kg N 2 H 4 ) #kg N 2 H 4 = (1)(70.9)(1)(1)(1000) = kg N 2 H 4

46 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = kg N 2 H 4 (b) actual yield (0.299 kg product) # kg N 2 H 4 =

47 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = kg N 2 H 4 (b) actual yield (0.299 kg product) (98.0 kg N 2 H 4 ) # kg N 2 H 4 = (100 kg product) purity factor

48 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = kg N 2 H 4 (b) actual yield (0.299 kg product) (98.0 kg N 2 H 4 ) # kg N 2 H 4 = (100 kg product) = kg N 2 H 4

49 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = kg N 2 H 4 (b) actual yield# kg N 2 H 4 = kg N 2 H 4

50 Dr. S. M. Condren EXAMPLE A chemical plant obtained kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH > N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = kg N 2 H 4 (b) actual yield# kg N 2 H 4 = kg N 2 H 4 (c) percent yield kg % yield = X 100 = 65.0 % yield 0.451kg


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