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Stoichiometry

Stoichiometry. Chapter 9 (12). Calculations based on a balanced chemical equation. Example using stoichiometry:. Why do I need to know about stoichiometry?. How many loaves of bread can be made from:. • 3 cups eggs. 2 loaves. Sample recipe for 1 loaf of bread: 8 cups flour 1 cup sugar

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Stoichiometry

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  1. Stoichiometry Chapter 9 (12) Calculations based on a balanced chemical equation

  2. Example using stoichiometry: Why do I need to know about stoichiometry? How many loaves of bread can be made from: • 3 cups eggs 2 loaves Sample recipe for 1 loaf of bread: 8 cups flour 1 cup sugar 2 cups milk 1.5 cups eggs 1/2 cup of butter 1/8 cup yeast • 1 cup milk 1/2 loaf • 6 cup flour 3/4 loaf

  3. 8 Fl + 1 S + 2 M + 1.5 E + 1/2 B + 1/8 Y => 1 Lf Sample recipe for 1 loaf of bread: 8 cups flour 1 cup sugar 2 cups milk 1.5 cups eggs 1/2 cup of butter 1/8 cup yeast “Chemical Equation” for bread recipe:

  4. 2 2 3 Review of Balancing Equations What do the coefficients mean? • Molecules KClO3 --> KCl + O2 “2 molecules KClO3 produces 3 molecules of O2”

  5. 2 2 3 Review of Balancing Equations What do the coefficients mean? • Molecules KClO3 --> KCl + O2 “2 molecules of KCl are formed when 3 molecules of O2 are formed”

  6. 2 2 3 Review of Balancing Equations What do the coefficients mean? b) Moles KClO3 => KCl + O2 “2 moles of KCl are formed when 3 moles of O2 are formed”

  7. 2 2 3 Review of Balancing Equations What do the coefficients mean? b) Moles KClO3 => KCl + O2 “2 moles of KCl are formed when 2 moles of KClO3 are decomposed”

  8. In the following reaction how many moles of PbCl2 are formed if 5.000 moles of NaCl react? 2 NaCl + Pb(NO3)2  PbCl2 + 2 NaNO3 1 moles PbCl2 = 2 moles NaCl 2.500 moles PbCl2 5.000 moles NaCl

  9. In the following reaction how many moles of NH3 are formed if 4.0 moles of H2 react? N2 + 3 H2 => 2 NH3 2 moles NH3 = 2.7 moles NH3 3 moles H2 4.0 moles H2 Complete Problems 1-5 on the practice page.

  10. In the following reaction how many grams of NH3 are formed if 4.00 moles of H2 react? N2 + 3 H2 => 2 NH3 2 moles NH3 17 g NH3 = 1 moles NH3 3 moles H2 45.3 g NH3 4.00 moles H2 coefficients Moles A Moles B 1 1 mw mw Grams B Grams A

  11. In the following reaction how many moles of NH3 are formed if 10.0 grams of H2 react? N2 + 3 H2 => 2 NH3 1 moles H2 2 mole NH3 = 3 moles H2 2.016 g H2 3.31 mol NH3 10.0 grams H2 coefficients Moles A Moles B 1 1 mw mw Grams B Grams A Complete Problems 6-10 on the practice page.

  12. In the following reaction how many grams of NH3 are formed if 25.0 grams of N2 react? N2 + 3 H2 => 2 NH3 1 moles N2 2 mole NH3 17 g NH3 = 1 moles N2 1 mole NH3 28.02 g N2 30.3 g NH3 25.0 g N2 coefficients Moles A Moles B 1 1 mw mw Grams B Grams A Complete Problems 10-15 on the practice page.

  13. How many grams of NH3 are formed if 25.0 grams of N2 react with 10.0 g of H2? N2 + 3 H2 => 2 NH3 25.0 g N2 (Solve the problem separately with each number) 1 moles N2 2 mole NH3 17 g NH3 = 1 moles N2 1 mole NH3 28.02 g N2 30.3 g NH3 10.0 grams H2 1 moles H2 2 mole NH3 17 g NH3 3 moles H2 1 mole NH3 2 g H2 56.7 g NH3 56.7 g NH3 = (The smaller answer is the only correct one)

  14. Complete problems 16-20.

  15. How many grams of NH3 are formed if 25.0 grams of N2 react with 10.0 g of H2? N2 + 3 H2 => 2 NH3 25.0 g N2 1 moles N2 2 mole NH3 17 g NH3 = 1 moles N2 1 mole NH3 28.02 g N2 30.3 g NH3 10.0 grams H2 1 moles H2 2 mole NH3 17 g NH3 3 moles H2 1 mole NH3 2 g H2 56.7 g NH3 56.7 g NH3 = How much of the excess reagent is left over?

  16. How many grams of H2 (the excess reagent) are required to react with 25.0 g of N2 (the limiting reagent) ? N2 + 3 H2 => 2 NH3 25.0 g N2 1 moles N2 3 mole H2 2 g H2 = 1 moles N2 1 mole H2 28 g N2 5.36 g H2 REQUIRED Left over = Given amount – Required amount = 10.0 g H2 - 5.36 g H2 = 4.64 g H2

  17. How many grams of NH3 are formed if 10.0 grams of N2 react with 15.0 g of H2? How much of the excess reagent is left over? N2 + 3 H2 => 2 NH3

  18. Percent Yield Calculations Terms: Theoretical Yield = the CALCULATED amount of product expected Actual Yield = the EXPERIMENTAL amount that was actually obtained Actual X 100 % Yield = Theoretical

  19. What is the percent yield in a reaction where 1.50 mol of NH3 was obtained after reacting 10.0 g of H2 with excess nitrogen? N2 + 3 H2 => 2 NH3 1 moles H2 2 mole NH3 = 3 moles H2 2.016 g H2 3.31 mol NH3 10.0 grams H2 Theoretical Yield 1.50 % yield = X 100 = 45.3% 3.31

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