1 / 19

Lecture 4.4 – Particles to Particles and Complex Stoichiometry

Lecture 4.4 – Particles to Particles and Complex Stoichiometry. Today’s Learning Targets. LT 4.7 – I can convert from particles of one compound to particles of another compound

jeneva
Télécharger la présentation

Lecture 4.4 – Particles to Particles and Complex 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. Lecture 4.4 – Particles to Particles and Complex Stoichiometry

  2. Today’s Learning Targets • LT 4.7 – I can convert from particles of one compound to particles of another compound • LT 4.8 – I can complete a complex stoichiometric conversion that incorporates molarity, particles, mass, moles, and volumes of substances.

  3. Molarity of Substance B Molarity of Substance A Moles of B Grams of Substance B Moles of A Grams of Substance A Atoms of Substance A Atoms of Substance B

  4. Our Focus for Today Moles of A Moles of B Atoms of Element or Compound B Atoms of Element or Compound A

  5. How do we convert particles of one substance to particles of another substance?

  6. I. Moles to Particles Conversion Factor • Our conversion factor for moles  particles is: 6.022 x 1023 particles mole Can be reversed to fit conversion

  7. II. Moles to Particles Between 2 Substances • We can utilize what we know about converting between moles to convert between particles for a different substance. • We simply follow the same rules and conversion factors

  8. Class Example • You react 30 x 1023 particles of HCl for the following reaction: HCl + MnO2 MnCl4 + H2O • How many molecules of MnCl4 did you use for this reaction?

  9. Example • You react 3 moles of MnO2 for the following reaction: HCl + MnO2 MnCl4 + H2O • How many molecules of HCl did you use for this reaction?

  10. Table Talk • For the following reaction: NH4SCN + FeCl3NH3(g) + HCl + Fe(SCN) 3 • If you react 4 moles of FeCl3, how many particles of NH3 would you produce?

  11. Table Talk • If you react 12 x 1023 particles of MnO2 with HCl in the following reaction: HCl + MnO2 MnCl4 + H2O • How many moles of MnCl4 did you use in this reaction?

  12. How do I complete complex stoichiometry problems?

  13. Molarity of Substance B Molarity of Substance A Moles of B Grams of Substance B Moles of A Grams of Substance A Atoms of Substance A Atoms of Substance B

  14. Class Example You wish to make mustard gas using the reaction: (C2OH5)2S + 2 HCl → (C2H4Cl)2S + 2 H2O If I add 52 g of HCl, then how many particles of H2O will I produce?

  15. Molarity of Substance B Molarity of Substance A Moles of B Grams of Substance B Moles of A Grams of Substance A Atoms of Substance A Atoms of Substance B

  16. Table Talk • You wish to make mustard gas using the reaction: (C2OH5)2S + 2 HCl → (C2H4Cl)2S + 2 H2O • If I add 143 g of (C2OH5)2S , then how many grams of (C2H4Cl)2S will I produce?

  17. Molarity of Substance B Molarity of Substance A Moles of B Grams of Substance B Moles of A Grams of Substance A Atoms of Substance A Atoms of Substance B

  18. Table Talk • You run the Haber – Bosh Reaction to make NH3: N2 + 3H2  2NH3 • If you begin with 15 x 1023 particles of H2, then how many grams of NH3 do you produce?

  19. Molarity of Substance B Molarity of Substance A Moles of B Grams of Substance B Moles of A Grams of Substance A Atoms of Substance A Atoms of Substance B

More Related