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Chapter 4

Chapter 4. Quantities of Reactants and Products. Balanced Chemical Equation. representation of a chemical reaction which uses coefficients (prefix numbers) to represent the relative amounts of reactants and products.

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Chapter 4

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  1. Chapter 4 Quantities of Reactants and Products

  2. Balanced Chemical Equation • representation of a chemical reaction which uses coefficients (prefix numbers) to represent the relative amounts of reactants and products

  3. EXAMPLE How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2? 2 H2 + O2 2 H2O (3.3 mol O2) (2 mol H2O) #mol H2O = = 6.6 mol H2O (1 mol O2) Mole ratio from balanced chemical equation

  4. Reaction of H2 and I2

  5. Types of Reactions • synthesis reactions or combination reactions • decomposition reactions • displacement reactions • exchange reactions

  6. Types of Chemical Reactions

  7. Synthesis or Combination Reactions Formation of a compound from simpler compounds or elements.

  8. Combination Reaction

  9. Decomposition Reactions Separation into constituents by chemical reaction.

  10. Decomposition Reactions

  11. Dynamite

  12. Electrolysis • decomposition caused by an electric current • anode • electrode where oxidation occurs • cathode • electrode where reduction occurs

  13. Electrolysis

  14. Displacement Reactions Reaction of a compound with an element to produce a new compound and release a different element

  15. Displacement Reactions

  16. Exchange Reactions Reaction where ion partners are exchanged

  17. Writing and BalancingChemical Equations • Write a word equation. • Convert word equation into formula equation. • Balance the formula equation by the use of prefixes (coefficients) to balance the number of each type of atom on the reactant and product sides of the equation.

  18. Example Hydrogen gas reacts with oxygen gas to produce water. Step 1. hydrogen + oxygen  water Step 2. H2 + O2 H2O Step 3. 2 H2 + O2 2 H2O

  19. Example Iron(III) oxide reacts with carbon monoxide to produce the iron oxide (Fe3O4) and carbon dioxide. iron(III) oxide + carbon monoxide  Fe3O4 + carbon dioxide Fe2O3 + CO  Fe3O4 + CO2 3 Fe2O3 + CO  2 Fe3O4 + CO2

  20. Stoichiometry stoi·chi·om·e·trynoun Calculations of the quantitative relationships between reactants and products in a chemical reaction.

  21. The Mole and Chemical Reactions:The Macro-Nano Connection 2 H2 + O2 2 H2O 2 H2 molecules 1 O2 molecule 2 H2O molecules 2 moles H2 molecules 1 mole O2 molecules 2 moles H2O molecules 2 kmoles H2 molecules 1 kmole O2 molecules 2 kmoles H2O molecules 2 mmoles H2 molecules 1 mmole O2 molecules 2 mmoles H2O molecules 4 g H2 32 g O2 36 g H2O

  22. Stoichiometric Relationships

  23. 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. weld Photo by Mike Condren

  24. EXAMPLEHow 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  103 g/in The mass of iron in a weld adding 10% mass: #g = (1.67  103 g) (0.10) = 167 g

  25. EXAMPLEHow 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 = (1.67  103 g) (0.10) = 167 g Balanced chemical equation: Fe2O3 + 2 Al  2 Fe + Al2O3 What mass of Fe2O3 is required for the thermite process? (1 mol Fe) (1 mol Fe2O3) (159.7 g Fe2O3) #g Fe2O3 = (167 g Fe) (55.85 g Fe) (2 mol Fe) (1 mol Fe2O3) = 238 g Fe2O3

  26. EXAMPLEHow 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 Balanced chemical equation: Fe2O3 + 2 Al  2 Fe + Al2O3 What mass of Fe2O3 is required for the thermite process? #g Fe2O3 = 238 g Fe2O3 What mass of Al is required for the thermite process? (1 mol Fe) (2 mol Al) (26.9815 g Al) #g Al = (167 g Fe)  (55.85 g Fe) (2 mol Fe) (1 mol Al) = 80.6 g Al

  27. Limiting Reactant reactant that limits the amount of product that can be produced

  28. Limiting Reactant

  29. EXAMPLEWhat is the number of moles of CaSO4 (S) that can be produced by allowing 1.0 mol SO2, 2.0 mol CaCO3, and 3.0 mol O2 to react? 2SO2(g) + 2CaCO3(s) + O2(g) 2CaSO4(S) + 2CO2(g) balanced equation relates: 2SO2(g) 2CaCO3(s) O2(g) have only: 1SO2(g)2CaCO3(s)3O2(g) not enough SO2 to use all of the CaCO3 or the O2 not enough CaCO3 to use of the O2 SO2 is the limiting reactant

  30. EXAMPLEWhat is the number of moles of CaSO4 (S) that can be produced by allowing 1.0 mol SO2, 2.0 mol CaCO3, and 3.0 mol O2 to react? 2SO2(g) + 2CaCO3(s) + O2(g) 2CaSO4(S) + 2CO2(g) have only: 1SO2(g)2CaCO3(s)3O2(g) SO2 is the limiting reactant if use all of SO2 #CaCO3 = (1 mol SO2)(2 mol CaSO4/2 mol SO2) = 1 mol CaSO4

  31. Theoretical Yield the amount of product produced by a reaction based on the amount of the limiting reactant

  32. Actual Yield amount of product actually produced in a reaction

  33. Percent Yield actual yield % yield =  100 theoretical yield

  34. EXAMPLEA rocket fuel, hydrazine, is produced by a reaction ofCl2 that is reacted with excess NaOH and NH3. What (a) theoretical yield can be produced from 1.00 kg of Cl2? 2NaOH + Cl2 + 2NH3N2H4 + 2NaCl + 2H2O (a) to calculate the theoretical yield, use the net equation for the overall process (1 kmol Cl2) (70.9 kg Cl2) molar mass (1.00 kg Cl2) #kg N2H4 = (1 kmol N2H4) (1 kmol Cl2) balanced equation (32.0 g N2H4) (1 mol N2H4) molar mass = 0.451 kg N2H4

  35. EXAMPLE (b) What is the actual yield if 0.299 kg of 98.0% N2H4 is produced for every 1.00 kg of Cl2? 2NaOH + Cl2 + 2NH3 N2H4 + 2NaCl + 2H2O (a) theoretical yield #kg N2H4 = 0.451 kg N2H4 (b) actual yield (0.299 kg product) # kg N2H4 = (98.0 kg N2H4) (100 kg product) purity factor = 0.293 kg N2H4

  36. EXAMPLE (c) What is the percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 N2H4 + 2NaCl + 2H2O (a) theoretical yield #kg N2H4 = 0.451 kg N2H4 (b) actual yield # kg N2H4 = 0.293 kg N2H4 (c) percent yield 0.293 kg % yield =  100 = 65.0 % yield 0.451kg

  37. Combustion Analysis

  38. ExampleBenzoic acid is known to contain only C, H, and O. A 6.49-mg sample of benzoic acid was burned completely in a C-H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO2 and 2.85-mg of H2O formed. What is the empirical formula of benzoic acid? (16.4-mg of CO2 )(12.01-mg C) #mg C = = 4.48-mg C (44.01-mg CO2) 4.48-mg C %C =  100 = 68.9% C 6.49-mg sample

  39. ExampleBenzoic acid is known to contain only C, H, and O. A 6.49-mg sample of benzoic acid was burned completely in a C-H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO2 and 2.85-mg of H2O formed. What is the empirical formula of benzoic acid? (2.85-mg of H2O)(2.02-mg H) #mg H = = 0.319-mg H (18.02-mg H2O) 0.319-mg H %C =  100 = 4.92% H 6.49-mg sample

  40. ExampleBenzoic acid is known to contain only C, H, and O. A 6.49-mg sample of benzoic acid was burned completely in a C-H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO2 and 2.85-mg of H2O formed. What is the empirical formula of benzoic acid? 68.9% C4.92% H % O = (100 - (68.9% C + 4.92% H) = 26.2% O

  41. ExampleBenzoic acid is known to contain only C, H, and O. A 6.49-mg sample of benzoic acid was burned completely in a C-H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO2 and 2.85-mg of H2O formed. What is the empirical formula of benzoic acid? % C 68.9 H 4.92 O 26.2 Relative # Atoms (%/gaw) 68.9/12.0 = 5.75 4.92/1.01 = 4.87 26.2/16.0 = 1.64 Divide by Smallest 5.75/1.64 = 3.51 4.87/1.64 = 2.97 1.64/1.64 = 1.00 Multiply by Integer 3.51  2 = 7 2.97  2 = 6 1.00  2 = 2 C7H6O2

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