Intro to Ch 9

1. Intro to Ch 9 • Pg 267 #2= work w/partner (a-f)=10 min • How many eggs do you need? 3 eggs b. If you use all the butter (and get enough eggs), how many cookies can you make? 4 dozen cookies c. How much butter is needed to react with all the eggs? 1.2 cups butter (per 1.8 eggs)

2. d. How many cookies can you make? 3.6 dozen e. Which will you have left over, eggs or butter? butter f. How much is left over? 0.133cups

3. g. Relate this question to the concepts of chemical stoichiometry. -balanced equation is like a well-written recipe -you can make more or less by multiplying coefficients by an n-value -an unbalanced equation is like a recipe that lists ingredients only, not quantities = YUCK! = ALWAYS BALANCE FIRST!!!!!!!!

4. Review top 248- What a balanced equation tells us

5. ALWAYS Balance First! • Because atoms are rearranged (not created or destroyed) in a chemical reaction, we must balance. • Coefficients- give relative number of molecules

6. Ch. 9 Notes -- Stoichiometry balanced • Stoichiometry refers to the calculations of chemical quantities from __________________ chemical equations. Interpreting Everyday Equations 2 guards + 2 forwards + 1 center  1 basketball team ___ + ___ + ___  __________ PracticeProblems: 1) How many guards does it take to make 7 teams? ______ 2) How many forwards are there in 8 teams? ______ 3) If you have 8 centers, 17 forwards and 14 guards, how many teams can be made? ______ What do you run out of first? _________ 2G 2F C G2F2C 14 16 7 guards

7. Interpreting Chemical Equations • ___N2(g) + ___H2(g) ___NH3(g) • The first thing that must be done is to ______________ the equation! • Here are the kinds of information you can get from the equation: • ____ mole N2 + ____ moles H2 ____ moles NH3 • ____ molecule N2 + ____ molecules H2 ____ molecules NH3 • ____ liter N2 + ____ liters H2 ____ liters NH3 • ____ grams N2 + ____ grams H2 ____ grams NH3 1 3 2 balance 1 3 2 1 3 2 1 3 2 28.0 6.0 34.0 To be discussed in later chapters

8. Stoichiometry 9.2 - Mole-Mole Conversions *Molar ratio- relates moles of 1 substance to moles of another in a balanced chem. equation*mole conversion factor comes from the coefficients of the balanced chemical equation. • Step 1: Write down the equation (if not already given) • Step 2: Predict the products (if not already given) • Step 3: BALANCE the equation (if not already done so) • Step 4: Write down the “given”. • Step5: Set up a conversion factor to change from moles to moles.

9. Ex: mol ratio Decomposition of water: • Write balanced chemical equation 2 H20(l) 2H2(g) + O2(g) For every __ moles H2O decomposed, __ mole O2 is produced (molar ratio)

10. 2 H20(l) 2H2(g) + O2(g) What if we created 2 mol O2 , predict the amount of H2O we started with. = 4mol H2O What if we had 5.8 mol water, how much H2 and O2 5.8mol H20  5.8 mol H2 + 2.9 mol O2

11. Self-check 9.1 pg 251 Calculate the moles of CO2 formed when 4.30 mol of propane, C3H8 , reacts with the required 21.5mol of O2 (hint: combustion) C3H8 (g) + 5O2 (g)  3CO2 (g) + 4 H2O (l)

12. Practice Problems:N2(g) + 3H2 (g) 2NH3(g) • How many moles of ammonia can be made from 7 moles of nitrogen reacting with an excess of hydrogen? • 2) How many moles of hydrogen are required to completely react with 8 moles of nitrogen to produce ammonia? • 3) How many moles of hydrogen are needed to react with an excess of nitrogen to make 10 moles of ammonia? 2 mol NH3 7 mol N2 = 14 moles of NH3 x 1 mol N2 3 mol H2 8 mol N2 x = 24 moles of H2 1 mol N2 3 mol H2 10 mol NH3 = 15 moles of H2 x 2 mol NH3

13. Other Conversion Problems • Mass-Mass: (grams to moles to moles to grams) • Step 1: Write down the equation (if not already given) • Step 2: Predict the products (if not already given) • Step 3: BALANCE the equation (if not already done so) • Step 4:Write down the “given” and convert from grams to moles. • Step5: Convert from moles of the “given” to moles of the “unknown” using a mole conversion factor. • Step 6: Convert from moles of the unknown to grams.

14. N2(g) + 3H2(g) 2NH3(g) Practice Problem: How many grams of ammonia can be made from reacting 39.0 grams of nitrogen with an excess of hydrogen? 1 mol N2 2 mol NH3 17.0 g NH3 39.0 g N2 47.4 g NH3 x x x = 28.0 g N2 1 mol N2 1 mol NH3

15. Limiting Reagent (or Limiting Reactant) runs out • The limiting reagent is the reactant that ___________ _____ first. • The reactant that is in abundance is called the ___________ reagent. excess

16. Example Problems • . Consider the equation: A + 5B  3C + 4D. When equal mols of A and B are reacted, which is limiting? • a. A is the limiting reactant. • b. B is the limiting reactant • c. Neither are the limiting reactan • d. Need more information to solve • Answer: is the limiting reactant

17. Calculations Step 1: Do a mole-mole conversion for one of the substances. This answer is how much of it you need. Step 2: Compare your answer to how much reactant was given. Do you have enough? If not, this reactant is your limiting reagent! Practice Problems:N2(g) + 3H2(g) 2NH3(g) 1) If 2.7 moles of nitrogen reacts with 6.3 moles of hydrogen, which will you run out of first? 2) If 3.9 moles of nitrogen reacts with 12.1 moles of hydrogen, what is the limiting reagent? Were you given enough H2? ______ Limiting Reagent = ______ 3 mol H2 No ! (6.3<8.1) 2.7 mol N2 8.1 moles of H2 are needed for the reaction. x = 1 mol N2 H2 Were you given enough H2? ______ Limiting Reagent = ______ 3 mol H2 Yes ! (12.1>11.7) 3.9 mol N2 11.7 moles of H2 are needed for the reaction. x = 1 mol N2 N2

18. Excess Reagent (or Excess Reactant) • How many moles of excess reagent do you have? • Step 1: Do a mole-mole conversion starting with the limiting reagent as the given. • The answer you get is how much of the excess reagent you need to completely react with the limiting reagent. • Sometimes you get lucky and you already did this conversion from the previous problem! • Step 2: Subtract this answer from the amount given in the original problem, and that is how many moles of excess reagent there are.

19. Excess Reagent (or Excess Reactant) Practice Problem: Find the number of moles of excess reagent from the previous practice problems. 1 mol N2 6.3 mol H2 2.1 moles of N2 are needed for the reaction. x = 3 mol H2 2.7 moles of N2 were originally given, so the excess will be… 2.7 moles given − 2.1 moles needed = 0.6 moles of N2 excess • For the second practice problem, we already started with the limiting reagent, so all you have to do is subtract… 12.1 moles given −11.7 moles needed= 0.4 moles of H2 excess

20. Percent Yield • Percent Yield is a ratio that tells us how ________________ a chemical reaction is. • The higher the % yield, the more efficient the reaction is. • Actual or “experimental” Yield • Theoretical or “ideal” Yield • The ___________ yield is the amount you experimentally get when you run the reaction in a lab. • The _______________ yield is the amount you are ideally supposed to get if everything goes perfectly. You can calculate this amount using stoichiometry! efficient x 100 % Yield = actual theoretical

21. Percent Yield PracticeProblem:2H2(g) + O2(g) 2H2O (g) 1) A student reacts 40. grams of hydrogen with an excess of oxygen and produces 300. grams of water. Find the % yield for this reaction. Step 1: Do a mass-mass conversion starting with the given reactant and converting to the product, (in this example, the water.) The answer you get is how much water you “theoretically” should have produced. Step 2: The other value in the question, (300 grams), is what you actually produced. Divide them to get your % yield! 1 mol H2 2 mol H2O 18.0 g H2O 40. g H2 360 g H2O x x x = 2.0 g H2 2 mol H2 1 mol H2O 300. % Yield = x 100 = 83.% 360