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

2. Why do I need to know about stoichiometry? 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 How many loaves of bread can be made from: 3 cups eggs2 loaves 1 cup milk 1/2 loaf 6 cup flour 3/4 loaf Example using stoichiometry:

3. 8 Fl + 1 S + 2 M + 1.5 E + 1/2 B + 1/8 Y => 1 Lf “Chemical Equation” for bread recipe: 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

4. Review of Balancing Equations KClO 3 --> KCl + O 2 322 What do the coefficients mean? a)Molecules “2 molecules KClO 3 produces 3 molecules of O 2 ”

5. Review of Balancing Equations KClO 3 --> KCl + O 2 322 What do the coefficients mean? a)Molecules “2 molecules of KCl are formed when 3 molecules of O 2 are formed”

6. Review of Balancing Equations KClO 3 => KCl + O 2 322 What do the coefficients mean? b) Moles “2 moles of KCl are formed when 3 moles of O 2 are formed”

7. Review of Balancing Equations KClO 3 => KCl + O 2 322 What do the coefficients mean? b) Moles “2 moles of KCl are formed when 2 moles of KClO 3 are decomposed”

8. In the following reaction how many moles of PbCl 2 are formed if 5.000 moles of NaCl react? 2 NaCl + Pb(NO 3 ) 2  PbCl 2 + 2 NaNO 3 5.000 moles NaCl 2 moles NaCl 1 moles PbCl 2 = 2.500 moles PbCl 2

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

10. In the following reaction how many grams of NH 3 are formed if 4.00 moles of H 2 react? N 2 + 3 H 2 => 2 NH 3 4.00 moles H 2 3 moles H 2 2 moles NH 3 = 45.3 g NH 3 1 moles NH 3 17 g NH 3 Grams A Moles AMoles B Grams B 11 coefficients mw

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

12. In the following reaction how many grams of NH 3 are formed if 25.0 grams of N 2 react? N 2 + 3 H 2 => 2 NH 3 25.0 g N 2 28.02 g N 2 1 moles N 2 = 30.3 g NH 3 1 moles N 2 2 mole NH 3 Grams A Moles AMoles B Grams B 11 coefficients mw 1 mole NH 3 17 g NH 3 Complete Problems 10-15 on the practice page.

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

14. Complete problems 16-20.

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

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

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

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

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