1 / 8

Stoichiometry

Stoichiometry. Preparation for College Chemistry Luis Avila Columbia University Department of Chemistry. The Mole-ratio method. Mole-Mole Calculations. Mole-Mass Calculatios. Mass-Mass calculations. Yield calculations. 1 molKMnO 4. 1 molKCl.

johneverett
Download Presentation

Stoichiometry

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. Stoichiometry Preparation for College Chemistry Luis Avila Columbia University Department of Chemistry

  2. The Mole-ratio method Mole-Mole Calculations Mole-Mass Calculatios Mass-Mass calculations Yield calculations

  3. 1 molKMnO4 1 molKCl KMnO4 + HCl + H2S KCl + MnCl2 + S + H2O 2 mol KMnO4 2mol KMnO4 2 mol KMnO4 8 mol H2O 5 mol S 5 mol H2S 5 moleS 2 mole KMnO4 Mole-Ratio 2 6 5 2 2 5 8 How many moles of S can be obtained from 1.5 mole KMnO4 ? 1.5 mole KMnO4 x = 3.8 Mole S

  4. KMnO4 + HCl + H2S KCl + MnCl2 + S + H2O 1 mol KMnO4 5 mol S 32.07 g S 158.04 g KMnO4 2mol KMnO4 1mol S Mole-Mass Calculation 2 6 5 2 2 5 8 How many g of S can be obtained from 1.5 g KMnO4 ? 1.5 g KMnO4 x x x = 80.9 g S

  5. Limiting Reactant

  6. 1 mol Fe 1 mol Fe3O4 55.85 g Fe 3 mol Fe 3Fe(s) + 4H2O(g) Fe3O4 + 4H2 1 mol H2O 1 mol Fe3O4 18.02 g H2O 4 mol H2O Limiting Reactant Calculation 16.8 g 10.0 g 1. Calculate amount of product formed by each reactant .100 mol Fe3O4 .100 mol Fe3O4 16.8 g Fe 16.8 g Fe x x = yield of product .139 mol Fe3O4 10.0 g H2O x x = 2. The Limiting reactant gives the least amount of product.

  7. 16.8 g 10.0 g 1 mol Fe 55.85 g Fe 4 mol H2O 18.02 g H2O 3 mol Fe 1 mol H2O 3Fe(s) + 4H2O(g) Fe3O4 + 4H2 Excess Reactant Calculation 3. Calculate the excess amount by subtracting the reacted amount from the starting quantity 16.8 g Fe x x x = 7.23g H2O Excess water: 10.0 g - 7.2 g = 2.7 g unreacted water

  8. Actual Yield x 100 Theoretical Yield 231.55 g Fe3O4 1 mol Fe3O4 16.2 g Fe3O4 23.2 g Fe3O4 Percentage Yield Calculation In the previous reaction the theoretical yield was 23.2 g Fe3O4 .100 mol Fe3O4 x = If the actual amount obtained is 16.2 g, then the % yield: Percentage Yield = x 100 = 69.8%

More Related