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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

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Lecture 4.4 – Particles to Particles and Complex Stoichiometry

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  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

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