1 / 24

Free Group Study Tutoring

Need help with stoichiometry calculations? Join our free group study tutoring sessions every Monday, Tuesday, and Friday. Tutoring begins on Friday, Sept. 16th at 11am in the Learning Assistance Center. For more information, call 562-985-5350.

amynelson
Download Presentation

Free Group Study Tutoring

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Free Group Study Tutoring • Monday 5-6pm • Tuesday 5-6pm • Friday 11am-12pm • Tutoring will begin Friday Sept. 16th at 11am in the Learning Assistance Center! • For more information call the LAC at: 562-985-5350

  2. Stoichiometry stoi·chi·om·e·trynoun Literally – measuring the components Calculations of the quantitative relationships between reactants and products in a chemical reaction.

  3. 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 4.0 g H2 32.0 g O2 36.0 g H2O

  4. Stoichiometric Relationships

  5. 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 Fe/yard and that the weld is 1/10 inch wide. weld Photo by Mike Condren

  6. The mass of iron in 1/10 inch of this rail is: Fe2O3 + 2 Al  Al2O3 +2 Fe

  7. 130 9 0 0 Compound XCl4 contains 74.76% Cl by mass. Which element is X? • C • Ge • Si • Sn • Ti

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

  9. Limiting Reactant

  10. 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 all of the O2 SO2 is the limiting reactant

  11. Compare Amounts of Product SO2 limits the amount of product formed

  12. Mass of Product • 95.0 g of chlorine and 27.0 g of phosphorus react to form PCl3. What mass of PCl3 is formed? P4 (s) + 6 Cl2 (g)  4 PCl3 (l)

  13. 27.0g phosphorus gives the smaller yield. Phosphorus is the limiting reactant it limits the amount of product formed

  14. Excess Reactant • We can also calculate the amount of the non-limiting reactant that is used up, and thus how much is left over 92.72 g Cl2 are required to react with all the P4Thus 95.0-92.72=2.3 g Cl2 are left over as excess

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

  16. Actual Yield The amount of product actually produced in a reaction

  17. Percent Yield actual yield % yield =  100% theoretical yield For example, if only 103.5 g of PCl3 were actually produced

  18. EXAMPLEA rocket fuel, hydrazine, is produced by a reaction ofCl2 with excess NaOH and NH3. (a) What 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) (1 kmol N2H4) (1 kmol Cl2) balanced equation (32.0 kg N2H4) (1 kmol N2H4) molar mass = 0.451 kg N2H4

  19. 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) (98.0 kg N2H4) (100 kg product) purity factor = 0.293 kg N2H4

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

  21. 130 10 0 0 When 1.0 mole of Fe(s) reacts with excess O2(g), 0.325 moles of Fe2O3 is produced. What is the percent yield of Fe2O3? • 16 % • 32 % • 50 % • 65 %

  22. Combustion Analysis

  23. 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) (2.85-mg of H2O )(2.01-mg H) # mg H = = 0.318-mg H (18.01-mg H2O)

  24. # mg O = 6.49 - 4.48 - 0.318 = 1.69 mg O Empirical Formula C7H6O2

More Related