1 / 13

Types of Chemical Reactions and Solution Stoichiometry

Types of Chemical Reactions and Solution Stoichiometry. Molarity. The concentration of a solution measured in moles of solute per liter of solution. mol = M L. Preparation of Molar Solutions.

talmai
Download Presentation

Types of Chemical Reactions and Solution 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. Types of Chemical Reactionsand Solution Stoichiometry

  2. Molarity The concentration of a solution measured in moles of solute per liter of solution. mol= M L

  3. Preparation of Molar Solutions Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution? • Step #1: Ask “How Much?” (What volume to prepare?) • Step #2: Ask “How Strong?” (What molarity?) • Step #3: Ask “What mass should I use?.” (Molar mass is?) 1.500 L 0.500 mol 58.44 g = 43.8 g 1 L 1 mol

  4. Dilution Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution? MstockVstock = MdiluteVdilute (11.6 M)(x Liters) = (3.0 M)(0.250 Liters) x Liters = (3.0 M)(0.250 Liters) 11.6 M = 0.065 L

  5. Dilution • It is not practical to keep solutions of many different concentrations on hand, so chemists prepare more dilute solutions from a more concentrated “stock” solution.

  6. Dilution • What would the final volume of a solution be if 50.0 ml of 0.500 M NaOH is diluted to 0.125 M? • 200 ml

  7. Dilution • 300.0 ml of NaOH has a molarity of 1.25 M. The original volume was 250.0 ml. What was the original molarity? • 1.5 M

  8. Dilution • 30.0 ml of 1.25 M KCl is diluted to 0.450 M. What volume of water was added to the solution? • 53 ml

  9. LEARN THESE FOR QUIZ

  10. Precipitation (Double) Replacement Reactions The ions of two compounds exchange places in an aqueous solution to form two new compounds. AX + BY  AY + BX One of the compounds formed is usually a precipitate (an insoluble solid), an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

  11. Precipitation (Double) replacement forming a precipitate… Precipitation (Double) replacement (ionic) equation Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq) Complete ionic equation shows compounds as aqueous ions Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) +2 I-(aq)  PbI2(s) + 2K+(aq) + 2 NO3-(aq) Net ionic equation eliminates the spectator ions Pb2+(aq) + 2 I-(aq)  PbI2(s)

  12. Problem • Using your solubility rules, predict what will happen with the following pair of solutions are mixed: KNO3 (aq) and BaCL2 (aq). • Possible solid products? • K+ + Cl- and NO3- + Ba2+ • KCL and Ba(NO3)2 are soluble in water • Therefore, no precipitate forms

  13. Homework • Problems #35, 37

More Related