1. Ch. 9 Notes -- Stoichiometry Stoichiometry refers to the calculations of chemical quantities from __________________ chemical equations. Interpreting Everyday Equations 2 guards + 2 forwards + 1 center  1 basketball team ___ + ___ + ___  __________ Practice Problems: 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? _________ balanced 2G2FCG2F2CG2F2C 14 16 7guards

2. Interpreting Chemical Equations ___N 2 (g) + ___H 2 (g)  ___NH 3 (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 N 2 + ____ moles H 2  ____ moles NH 3 ____ molecule N 2 + ____ molecules H 2  ____ molecules NH 3 ____ liter N 2 + ____ liters H 2  ____ liters NH 3 ____ grams N 2 + ____ grams H 2  ____ grams NH 3 32 balance 132 132 132 28.06.034.0 1 To be discussed in later chapters

3. Stoichiometry Map Atoms/Particles/Molecules 6.022 x 10 23 Mass (always in grams) Periodic table Atoms/Particles/Molecules 6.022 x 10 23 Mass (always in grams) Periodic table Mole Substance B Number 1 Mole Substance A Number 1 Must use MOLAR ratio (the coefficients) from BALANCED chemical equation from Substance A to Substance B

4. Stoichemertry Mole-Mole Conversions 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.

5. Practice Problems: N 2 (g) + 3H 2 (g)  2NH 3 (g) 1) How many moles of ammonia can be made from 7.0 moles of nitrogen reacting with an excess of hydrogen? 2) How many moles of hydrogen are required to completely react with 8.0 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? 1 mol N 2 2 mol NH 3 7.0 mol N 2 x = 14 moles of NH 3 1 mol N 2 3 mol H 2 8.0 mol N 2 x= 24 moles of H 2 2 mol NH 3 3 mol H 2 10. mol NH 3 x = 15 moles of H 2

6. 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.

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

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

9. 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: N 2 (g) + 3H 2 (g)  2NH 3 (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? 1 mol N 2 3 mol H 2 2.7 mol N 2 x= 8.1 moles of H 2 are needed for the reaction. Were you given enough H 2 ? ______ Limiting Reagent = ______ No ! (6.3<8.1) H2H2 1 mol N 2 3 mol H 2 3.9 mol N 2 x= 11.7 moles of H 2 are needed for the reaction. Were you given enough H 2 ? ______ Limiting Reagent = ______ Yes ! (12.1>11.7) N2N2

10. 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. Excess Reagent (or Excess Reactant)

11. Practice Problem: Find the number of moles of excess reagent from the previous practice problems. Excess Reagent (or Excess Reactant) 3 mol H 2 1 mol N 2 6.3 mol H 2 x= 2.1 moles of N 2 are needed for the reaction. 2.7 moles of N 2 were originally given, so the excess will be… 2.7 moles given − 2.1 moles needed = 0.6 moles of N 2 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 H 2 excess

12. 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! Percent Yield x 100 % Yield = efficient actual theoretical

13. Practice Problem: 2H 2 (g) + O 2 (g)  2H 2 O (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! Percent Yield 2.0 g H 2 2 mol H 2 O1 mol H 2 40 g H 2 x x = 360 g H 2 O 2 mol H 2 x 1 mol H 2 O 18.0 g H 2 O % Yield = 300 360 x 100 =83.3%