1 / 22

Metal + Acid Displacement

Metal + Acid Displacement. Activity Series of Metals. Activity Series of Metals. metals higher in series react with compounds of those below metals become less reactive to water top to bottom metals become less able to displace H 2 from acids top to bottom. Potassium + Water.

jeremywilliam
Télécharger la présentation

Metal + Acid Displacement

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. Metal + Acid Displacement

  2. Activity Series of Metals

  3. Activity Series of Metals • metals higher in series react with compounds of those below • metals become less reactive to water top to bottom • metals become less able to displace H2 from acids top to bottom

  4. Potassium + Water

  5. Activity Series of Metals Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s) Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s) Fe(s) + 2HCl(aq) FeCl2(aq) + H2(g) Zn(s) + 2HBr(aq) ZnBr2(aq) + H2(g)

  6. Metal + Metal Salt Displacement

  7. 130 10 0 0 Which of the following reactions does NOT happen? • Cu(s)+H2SO4(aq)CuSO4(aq)+H2(g) • 2HNO3(aq)+2K(s)2KNO3(aq)+H2(g) • FeCl2(aq)+Zn(s)ZnCl2(s)+Fe(s) • Ca(s)+2H2O(l)Ca(OH)2(aq)+H2(g) • Cu(s)+2AgNO3(aq)2Ag(s)+Cu(NO3)2(aq)

  8. Solution A homogeneous mixture of two or more substances comprising the solvent which is the majority of the mixture and one or more solutes which are the smaller fraction. Concentration – how much solute is present in a given amount of solution

  9. Cola Drinks Solvent • water Solutes • carbon dioxide (gas) • sweetener (solid) • phosphoric acid (liquid) • caramel color (solid)

  10. Molarity – a measure of concentration The number of moles of solute per liter of solution. molarity  M moles of solute M = liters of solution units  molar = mol/L = M

  11. Preparation of 1.00 L of 0.0100 M KMnO4 solution from solid

  12. 130 10 0 0 Which value do you NOT need to determine the molarity of a solution? • Mass of solute • Molar mass of solute • Volume of solvent added • Total volume of solution

  13. Solution Preparation by Dilution

  14. Dilution • Molarity (mol/L) × Volume (L) = Moles • If we take a sample (say 25.0 mL) from a solution (say 0.372 M) and add extra water (say to a total volume of 500. mL) the moles of solute are unchanged • Thus M1V1 = moles = M2V2 • 0.372 M × 25.0 mL = M2 × 500. mL • M2 = 0.0186 M

  15. 130 10 0 0 How many L of conc HNO3 (16.0 M) are needed to prepare 0.500 L of 0.250 M nitric acid? • 32.0 L • 16.0 L • 0.500 L • 0.250 L • 0.00781 L

  16. Stoichiometric Relationships

  17. Titrationsat equivalence mol H+ = mol OH-

  18. EXAMPLE: A sample of lye, sodium hydroxide, is neutralized by sulfuric acid. How many milliliters of 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH? 2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O (25.0 mL NaOH) #mL H2SO4 = (1 L) (1000 mL) (0.400 mol NaOH) (1 L NaOH) (1000 mL) (1 L) (1 L H2SO4) (0.200 mol H2SO4) (1 mol H2SO4) (2 mol NaOH) = 25.0 mL H2SO4

  19. Ion Concentrations • 0.100 M NaCl • NaCl(aq) Na+(aq) + Cl-(aq) • Is 0.100 M in both Na+ and Cl- • 0.100 M Al2(SO4)3 • Al2(SO4)3(aq)  2 Al3+(aq) + 3 SO42-(aq) • Is 0.200 M in Al3+ and 0.300 M in SO42-

  20. Final Ion Concentrations • Consider the NaOH + H2SO4 reaction. What are the final concentrations of Na+ and SO42-? • Stoichiometric reaction, can use either reactant to determine moles of product • 25.0 mL NaOH × 0.400 mol NaOH × 1 L 1 L 1000 mL0.0100 mol NaOH × 1 mol Na2SO4 = 0.0050 mol Na2SO4 2 mol NaOH

  21. Total volume = 25.0 mL + 25.0 mL = 50.0 mL = 0.0500 L • [Na2SO4] = 0.00500 mol = 0.100 M 0.0500 L • [Na+] = 2 × 0.100 M = 0.200 M • [SO42-] = 0.100 M

  22. 130 10 0 0 Which solution has the highest concentration of SO42-? • 0.20 M CuSO4 • 0.15 M Na2SO4 • 0.070 M Fe2(SO4)3 • 0.10 M Ce(SO4)2

More Related