1 / 14

Chapter #9

Chapter #9. Stoichiometry. Chapter 9.1. Composition stoichiometry deals with the mass relationships of elements in compounds. Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction. Always starts with a balanced equation!.

callum-gill
Télécharger la présentation

Chapter #9

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. Chapter #9 Stoichiometry

  2. Chapter 9.1 • Composition stoichiometry deals with the mass relationships of elements in compounds. • Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction. Always starts with a balanced equation!

  3. Mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. • 2Al + 3H2SO4 Al2(SO4)3 + 3H2 • 2 mol Al 2 mol Al 2 mol Al 3 mol H2SO4 Al2(SO4)3 3H2 Then they can all be flipped over. Then do the second compound and the 3rd & 4th.

  4. Molar Mass is the mass, in grams, of one mole of a substance. • Al2O3 Al (26.981) x 2=53.962 g 0 (15.999) x 3= 47.997 g 101.959 g 1 mol or 101.959 g 101.959 g 1 mol

  5. Chapter 9.2 Mol to mol ratio from your balanced equation • Mole to Mole conversions • Given # of x what you want mols what you have Example: 2H2 + O2 2H2O How many moles of oxygen are required to react with 45.2 moles of hydrogen? 45.2 mol H2 X 1 mol O2 2 mol H2 Mol of what you want = 22.6 mol of O2 =

  6. Mol to Mass (grams) Mol to mol ratio from balancedequation • Given # x mol want x molar mass of mols mol have of what you want Example: 2H2 + O2 2H2O How many grams of oxygen are required to react with 45.2 moles of hydrogen? 45.2 mol H2 X 1 mol O2 X 31.998 g 2 mol H2 1 mol O2 Grams of what you want = 723.1548 g of O2 =

  7. 3. Mass (grams) to Moles Mol to mol ratio from balanced equation • Given # x 1 x mol want of grams molar mass mol have of given Example: 2H2 + O2 2H2O How many moles of oxygen are required to react with 63.56 grams of hydrogen? 63.56 g H2 X 1 mol H2 X 1 mol O2 2.014 g H2 2 mol H2 Moles of what you want = 15.78 mol of O2 =

  8. 4. Mass (g) to Mass (g) Mol to mol ratio = • Given # x 1 x mol want x molar mass of grams molar mass mol have of what you want of given Example: 2H2 + O2 2H2O How many grams of oxygen are required to react with 63.56 grams of hydrogen? 63.56 g H2 X 1 mol H2 X 1 mol O2 X 31.998g = 2.014 g H2 2 mol H2 504.91 g of O2

  9. Chapter 9.3 • Limiting reactant is the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction. • The substance that is not used up completely in a reaction is called the excess reactant.

  10. If you are given: 12 legs, 2 heads, 2 bodies, 3 eyes, 4 antenna, and 5 mouth parts. Which part is the limiting reactant and which is the excess? Cootie example: It takes 6 legs, 1 head, 1 body, 2 eyes, 2 antenna, and 1 mouth part to make a cootie. 6 legs + 1 head + 1body+ 2 eyes + 2 antenna + 1 mouth part 1cootie

  11. Silicon dioxide is usually unreactive but reacts with hydrogen fluoride according to the following reaction: • SiO2 + 4HF SiF4 + 2H20 • If 2 moles of HF react with 4.5 moles of SiO2 which is the limiting reactant? How much excess reactant? 2 moles HF X 1 mol SiO2 4 mol HF 4.5 mol SiO2 X 4 mol HF 1 mol SiO2 Answers are what you Need Given is what you have = 0.50 mol SiO2 = 18 mol HF

  12. Percent Yield Percent Yield = Theoretical yield is the amount in grams that you should have gotten math problem (moles to grams or grams to grams). Actual yield is the amount that you did get in lab. Actual yield Theoretical yield X 100

  13. Aluminum reacts with an excess of copper (II) sulfate. If 1.85 g of Al react with an excess of copper (II) sulfate, the actual yield of copper is3.70 g. Calculate the percent yield of Cu. 2Al + 3CuSO4 Al2(SO4)3 + 3Cu 1.85 g Al X 1 mol X 3 mol Cu X 63.546g 26.98g 2 mol Al 1 mol Cu Percent yield = actual 3.70 theoretical 6.536 = 6.536 g Cu = 56.6 % X 100 X 100

  14. “Cartoon”. January 9, 2007. http://www.washbac.org/images/farside.gif • “Cootie Toys”. January 10, 2007. http://static.flickr.com/48/124090469_d6d095d9f9_t.jpg • “Cootie parts”. January 10, 2007. http://shadeaux.hypermart.net/PICS/cootie2.jpg

More Related