2. Chapter 9 Rev

3. Chemical Change Changes the chemical composition of a compound Burns Odors Color Change Release or Absorbs Energy

4. Rearrangement of Atoms Set of Atoms in a Particular arrangement Same set of atoms in a Different arrangement

5. Energy in reactions Exothermic Releases energy Spontaneous reaction Endothermic Absorbs energy nonspontaneous

6. WHY REACTIONS OCCUR PARTICLES MUST COLLIDE FOR REACTION TO OCCUR

7. Chemical Equations Mass of Reactant = mass of Products

8. Tips For Balancing Delay balancing single elements and elements that are in more than one compound If polyatomic ions are the same on both sides of the equation treat them as a single element Balance from left to right; one element in a compound, then the next element in the same compound Thus 2 H 2 plus 1 O 2 yields 2H 2 O

9. Grams A 1 mol MW A Coefficient B Coefficient A MW B 1 mole Grams of A --> Moles of A ---> Moles of B ----> Grams of B

10. Stoichiometry Grams A MW A 1 B A MW B 1 Grams B Mole A Mole B Will it Ever end?

11. Solutions Solute - the substance being dissolved Solvent - the substance doing the dissolving

12. Molarity Molarity - ratio of the amount of substance, solute, dissolved in a volume of solution Molarity (M) = Moles of solute Volume of solution M = mole or mole vol in 1 dm^3 1 liter

13. Condition Changes Grams A 1 MW A MWB = g B 1 Vol B = MB 1 MB = Vol B Molarity A Vol A 1

14. Mole Map Grams A 1 MW A B A MWB = g B 1 Vol B = MB 1 MB = Vol B Molarity A Vol A 1

15. Help!! A B Map the Problem

16. Calculating Percent Yields Actual yields are given in the problem Found Experimentally Theoretical Yields are calculated from beginning values Actual yield Theoretical Yield X 100% = % Yield

17. Practice Problems 2NaOH + H 2 SO 4 ----> 2H 2 O + Na 2 SO 4 How many grams of sodium sulfate will be formed if you start with 200.0 grams of sodium hydroxide and excess sulfuric acid? 200.0 g NaOH 1 mol 40.0 g 1 2 142.0 g Na 2 SO 4 1 mol 355.3 g Na 2 SO 4

18. 25.0 grams of FeO is decomposed in the following reaction. Determine the amount of Fe produced Knowns 2FeO ----> 2Fe + O 2 Mass A = 25.0 g FeO MW A = 72.0 g FeO MW B = 56.0 g Fe B = 2/2 or 1/1 A Unknown g of Fe 19.4 g Fe 25.0 g FeO 1 Fe 1 FeO 1 Mol 72.0 g 56.0 g 1 mol

19. Determine the concentration of 10.7 g of NaCl dissolved in 50.0 ml of water 10.7 g 1 58 g 1.050 l =3.69M

20. Determine the new concentration when 150 ml of water is added to 50.0 ml of 0.25 M MgCl 2 solution 0.25 M 0.050l 1.200 l = 0.0625M

21. Limiting Factor The amount of reactant that is completely consumed in a chemical reaction. Excess reactant is the reactant that is left over at the end of a reaction

22. In the following equation, 10.0 grams of oxygen is reacted with 15.0 g of 1ron determine the amount of Iron (III) oxide produced 4Fe + 3O 2 ----> 2Fe 2 O 3 10.0 g O 2 1 32 g 2 3 160 g 1 mole = 33.3 g 15.0g Fe 1 56 g 2 4 160 g = 21.5 g