1. Stoichiometry The culmination of the Semester Objective: Use what you have learned all semester and apply these skills for solving Stoichiometry problems

2. Review When combining elements, how do you balance for charge? When combining elements, how do you balance for charge? Why do we balance chemical equations? Why do we balance chemical equations? What law dictates the balancing of equations? What law dictates the balancing of equations? How many atoms are in one mol? How many atoms are in one mol? How many liters are in 1 mol of a gas? How many liters are in 1 mol of a gas? The knowledge of the prior questions will ensure your success in Stoichiometry! The knowledge of the prior questions will ensure your success in Stoichiometry!

3. Stoichiometry The calculation of quantities…that’s all! The calculation of quantities…that’s all! Chemists use stoich to keep track of the amounts of product and reactant in a chemical reaction. Chemists use stoich to keep track of the amounts of product and reactant in a chemical reaction. Usually done by recording number of mols, grams, or representative particles Usually done by recording number of mols, grams, or representative particles

4. Interpreting Chem Equations Chemical equations can be interpreted in a number of different ways: Chemical equations can be interpreted in a number of different ways: # atoms # atoms # mols # mols # molecules # molecules # ions # ions Mass Mass Volume Volume

5. Interpret N 2 + 3H 2  2NH 3 How many atoms of each? How many atoms of each? How many Molecules? How many Molecules? Mass of Nitrogen and Hydrogen? Mass of Nitrogen and Hydrogen?

6. Mass Given Moles Given Moles Wanted Mass Wanted X 1mole molar mass X Mole ratio from balanced equation Wanted/Given X Molar mass of wanted Volume Wanted X 22.4 L/mole Volume Given X 1 mole/22.4L Particles Wanted (atoms, molecules, formula units) X 6.02x10 23 Particles Given (atoms, molecules, formula units) X 1 mole/ 6.02x10 23 The Mole Road Map

7. Chemical Calculations N 2 + 3H 2  2NH 3 Mole Ratios, what is the mol ratio in this equation? Mole Ratios, what is the mol ratio in this equation?

8. Mole-Mole Calculations How many moles of ammonia are produced when 0.60mol of Nitrogen reacts with Hydrogen? How many moles of ammonia are produced when 0.60mol of Nitrogen reacts with Hydrogen? N 2 + 3H 2  2NH 3 1. List your knowns: mol N 2 =.60mol 2. Unknown: mol NH 3 3. Mole ratio 1:3:2 2 mol NH 3 : 1mol N 2

9. Mole-Mole Calculations 2 mol NH 3 : 1mol N 2 Next we setup up a conversion to eliminate mol N 2 and be left with mol NH 3 0.60mol N 2 x 2mol NH 3 0.60mol N 2 x 2mol NH 3 = 1.2 NH 3 1 mol N 2 Is 1.2 to.60 a 2:1 ratio?

10. Mole-Mole Calculations 5C + 2SO 2  CS 2 + 4 CO 1. How many mols of CS 2 form when 2.7 mol C reacts? Known: 2.7 mol C Unknown: mol CS 2 Mole Ratio: 5:2:1:4 1 CS 2 : 5 C

11. Mole-Mole Calculations 1 CS 2 : 5 C Setup conversion 2.7mol C x 1 mol CS 2 2.7mol C x 1 mol CS 2 =.54 mol CS 2 5 mol C 5 mol C Is there a 1:5 ratio between these two #’s? 2.7 :.54

12. Demo: Stoichiometry of Hydrogen Production Zinc + Hydrochloric acid  _____+______ Zinc + Hydrochloric acid  _____+______ Write the balanced chemical equation: Write the balanced chemical equation: Diagram of apparatus Diagram of apparatus Beaker A BC Beaker A BC 1.2g Zn2.4g Zn4.8 g Zn 1.2g Zn2.4g Zn4.8 g Zn Show calculations from each of the following below Moles H 2 _________________ Moles HCl________________ Vol. H 2 __________________

13. Limiting Reagents The substance in a chemical reaction that runs out first. The substance in a chemical reaction that runs out first.

14. Limiting Reagents Determines how much of a product you can make. Determines how much of a product you can make. Once the limiting reagent is used up, there can be no more reaction. Once the limiting reagent is used up, there can be no more reaction.

15. Which is Limiting or in Excess? Copper reacts with sulfur to form copper(I) sulfide. 2Cu + S  Cu 2 S What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?

16. Steps to Solve Limiting Problems 1. List the knowns: 80g Cu 25g S 2. Find # moles each reactant: g Cu  mol Cu g Cu  mol Cu g S  mol S g S  mol S 3. Find mole ratio 4. Which substance is in excess? Which is limited?

17. Solve 80 g Cu x 1mol Cu = 1.26 mol Cu 80 g Cu x 1mol Cu = 1.26 mol Cu 63.5 g Cu 63.5 g Cu 25.0 g S x 1mol S =.779 mol S 25.0 g S x 1mol S =.779 mol S 32.1 g S 32.1 g S 1.26 mol Cu x 1mol S =.63 mol S 1.26 mol Cu x 1mol S =.63 mol S 2 mol Cu 2 mol Cu.63 mol sulfur is the amount needed to complete the reaction. Since we were given.779 mol S and only needed.63 mol S…the Sulfur was in excess and Cu was the L-reagent.

18. Calcium Oxide reacts with water in an exothermic reaction, to make calcium hydroxide and heat Calcium Oxide reacts with water in an exothermic reaction, to make calcium hydroxide and heat CaO + H 2 O  Ca(OH) 2 + 65.2 kJ heat CaO + H 2 O  Ca(OH) 2 + 65.2 kJ heat If 20.0g of CaO reacts with 10.0g of water, which is the limiting reagent? If 20.0g of CaO reacts with 10.0g of water, which is the limiting reagent? Limiting Example

19. Answer CaO + H 2 O  Ca(OH) 2 + 65.2 kJ heat CaO + H 2 O  Ca(OH) 2 + 65.2 kJ heat 20.0 g CaO x 1mol/56.08g =.357 mol CaO 20.0 g CaO x 1mol/56.08g =.357 mol CaO 10.0 g H2O x 1 mol/18.02g =.554 mol H 2 O 10.0 g H2O x 1 mol/18.02g =.554 mol H 2 O From the balanced equation CaO and H 2 O are in a 1/1 ratio. Our actual amounts are not in a 1/1 ratio. From the balanced equation CaO and H 2 O are in a 1/1 ratio. Our actual amounts are not in a 1/1 ratio. The CaO is limiting. It will determine the amount of the products formed and the amount of heat released. The CaO is limiting. It will determine the amount of the products formed and the amount of heat released.

20. How many grams of Ca(OH) 2 are produced? CaO was limiting 20.0 g CaO x 1mol =.357mol 56.08g 56.08g.357 moles CaO x 1 Ca(OH) 2 x 74.28 g = a(OH) 2.357 moles CaO x 1 Ca(OH) 2 x 74.28 g = 26.5 g Ca(OH) 2 1 CaO 1 mol Ca(OH) 2 1 CaO 1 mol Ca(OH) 2

21. How much energy? Multiply moles of limiting reagent by energy produced. Multiply moles of limiting reagent by energy produced. If 65.2 kj is released when 1 mole of Cao is reacted, and we reacted.357 moles then: If 65.2 kj is released when 1 mole of Cao is reacted, and we reacted.357 moles then: Energy released =.357 moles X 65.2kj = 23.3 kj Energy released =.357 moles X 65.2kj = 23.3 kj