# Ch 9 Notes Stoichiometry Stoichiometry refers to the

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Ch. 9 Notes -- Stoichiometry • Stoichiometry refers to the calculations of balanced chemical quantities from_________ chemical equations. Measuring elements • Greek for “____________” • When baking cookies, a recipe is usually used, telling the exact amount of each ingredient double • If you need more, you can ________or triple _______the amount • Thus, a recipe is much like a balanced equation

Interpreting Everyday Equations 2 guards + 2 forwards + 1 center 1 basketball team ___ + ___ 2 G 2 F 2 C + ___ C _____ Practice Problems: 14 1) How many guards does it take to make 7 teams? ______ 2) How many forwards are there in 8 teams? ______ 16 3) If you have 8 centers, 17 forwards and 14 guards, how many teams can be made? ______ What do you run out of first? _____ 7 guards

Interpreting Chemical Equations 1 2 (g) + ___H 3 2 (g) ___NH 2 ___N 3 (g) • The first thing that must be done is to _______ the balance equation! • Here are the kinds of information you can get from the equation: ____ 1 mole N 2 + ____ 3 moles H 2 ____ 2 moles NH 3 ____ 1 molecule N 2 + ____ 3 molecules H 2 ____ 2 molecules NH 3 1 liter N 2 + ____ 3 liters H 2 ____ 2 liters NH 3 ____ 28. 0 grams N 2 + ____ 6. 0 grams H 2 ____ 34. 0 grams NH 3 ____

Mole-Mole Conversions coefficients The mole conversion factor comes from the _________ of the balanced chemical equation. Step 1: Write down the “given”. This becomes “x” Step 2: Set up a conversion factor to change from moles of “x” to moles of “y”. Practice Problems: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 1) How many moles of ammonia can be made from 7 moles of nitrogen reacting with an excess of hydrogen? 7 mol N 2 x 2 mol NH 3 = 14 moles of NH 3 1 mol N 2 2) How many moles of hydrogen are required to completely react with 8 moles of nitrogen to produce ammonia? 8 mol N 2 x 3 mol H 2 = 24 moles of H 2 1 mol N 2 3) How many moles of hydrogen are needed to react with an excess of nitrogen to 3 mol H 2 make 10 moles of ammonia? = 15 moles of H 2 10 mol NH 3 x 2 mol NH 3

Mole to Mole conversions continued 4) How many moles of O 2 are produced when 3. 34 moles of Al 2 O 3 decompose? 2 Al 2 O 3 → 4 Al + 3 O 2 3 mol O 2 = 3. 34 mol Al 2 O 3 x 5. 01 moles of O 2 2 mol Al 2 O 3 5) If 3. 84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? 2 C 2 H 2 + 5 O 2 → 4 CO 2 + 2 H 2 O x 3. 84 mol C 2 H 2 5 mol O 2 2 mol C 2 H 2 = 9. 6 moles of O 2

Other Conversion Problems • Volume-Volume: (treat it exactly like a ______ mole-mole conversion) Step 1: Write down the “given”. Step 2: Convert from liters to liters using a mole conversion factor. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Practice Problem: How many liters of ammonia can be made from reacting 125 liters of hydrogen with an excess of nitrogen? 125 L H 2 x 2 L NH 3 3 L H 2 = 83. 3 L NH 3 …the long way to get the same answer… 125 L H 2 x 1 mol H 2 x 2 mol NH 3 x 22. 4 L NH 3 = 83. 3 L NH 3 22. 4 L H 2 3 mol H 2 1 mol NH 3

Molar Mass Review Practice Problems: (1) Calculate the molar mass of each compound. (Add up all the atomic masses for each atom from the Periodic Table. ) a) Ca. CO 3 1 Ca = 1 x 40. 0 = 40. 0 1 C = 1 x 12. 0 = 12. 0 3 O’s =3 x 16. 0 = 48. 0 b) (NH 4)2 SO 4 Add them up! 100 g c) C 3 H 6 O 3 C’s = 3 x 12. 0 = 36. 0 6 H’s = 6 x 1. 0 = 6. 0 1 O = 1 x 16. 0 = 16. 0 2 N’s = 2 x 14. 0 = 28. 0 8 H’s = 8 x 1. 0 = 8. 0 1 S = 1 x 32. 0 = 32. 0 4 O’s = 4 x 16. 0 = 64. 0 Add them up! 132. 0 g d) H 3 PO 4 Add them up! 58. 0 g 3 H’s = 3 x 1. 0 = 3. 0 1 P = 1 x 31. 0 = 31. 0 4 O’s = 4 x 16. 0 = 64. 0 Add them up! 98. 0 g

Stoichiometry Notes Part 2 • Mole-Mass: (moles to grams) Step 1: Write down the “given” (start moles of “x”). Step 2: Convert from moles of the “given” (x) to moles of the “unknown” (y) using a mole conversion factor. Step 3: Convert from moles of “y” to grams of “y” using molar mass. x N 2 (g) + y 3 H 2 (g) 2 NH 3 (g) Practice Problem: How many grams of ammonia, NH 3, can be made from reacting 20. 0 moles of hydrogen with an excess of nitrogen? 20. 0 moles H 2 x 2 mol NH 3 17. 0 g NH 3 x = 226. 67 g NH 3 3 mol H 2 1 mol NH 3

Other Conversion Problems • Mass-Mass: (grams to moles to grams) Step 1: Write down the “given” (x) and convert from grams to moles. Step 2: Convert from moles of the “given” (x) to moles of the “unknown” (y) using a mole conversion factor. Step 3: Convert from moles of “y” to grams of “y”. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Practice Problem: How many grams of ammonia can be made from reacting 39. 0 grams of nitrogen with an excess of hydrogen? 39. 0 g N 2 x 1 mol N 2 x 2 mol NH 3 x 17. 0 g NH 3 = 28. 0 g N 2 1 mol NH 3 47. 4 g NH 3

Other Conversion Problems • Mass-Mass: (grams to moles to grams) Step 1: Write down the “given” (x) and convert from grams to moles. Step 2: Convert from moles of the “given” (x) to moles of the “unknown” (y) using a mole conversion factor. Step 3: Convert from moles of “y” to grams of “y”. Cu + 2 Ag. NO 3 2 Ag + Cu(NO 3)2 Practice Problem: How many grams of silver are produced from the reaction of 17. 0 g of copper with an excess of silver nitrate, Ag. NO 3? 17. 0 g Cu x 1 mol Cu x 2 mol Ag x 108 g Ag 64 g Cu 1 mol Ag = 57. 38 g of Ag

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

Practice Problem: Percent Yield 2 H 2 (g) + O 2 (g) 2 H 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. 40 g H 2 x 1 mol H 2 x 2 mol H 2 O x 18. 0 g H 2 O = 360 g H 2 O 2. 0 g H 2 2 mol H 2 1 mol H 2 O Step 2: The other value in the question, (300 grams), is what you actually produced in the lab. Divide them to get your % yield! % Yield = 300 x 100 = 83. 3% 360

Percent Yield Practice Problem 7 H 202 + N 2 H 4 2 HNO 3 + 8 H 2 O 2) The reaction of 11 g of N 2 H 4 with excess H 2 O 2 according to the following reaction yields 35 g of HNO 3. What is the percent yield? Step 1: Do a mass-mass conversion starting with the given reactant and converting to the product, (in this example, the HNO 3. ) The answer you get is how much water you “theoretically” should have produced. 11 g N 2 H 4 x 1 mol N 2 H 4 32 g N 2 H 4 x 2 mol HNO 3 1 mol N 2 H 4 63 g HNO 3 x = 1 mol HNO 3 43. 3 g HNO 3 Step 2: The other value in the question, (35 grams), is what you actually produced. Divide them to get your % yield! % Yield = 35 x 100 = 80. 8% 43. 3

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

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) + 3 H 2 (g) 2 NH 3 (g) 1) If 2. 7 moles of nitrogen reacts with 6. 3 moles of hydrogen, which will you run out of first? Were you given enough 2. 7 mol N 2 x 3 mol H 2 1 mol N 2 = 8. 1 moles of H 2 are H 2? ______ No ! (6. 3<8. 1) needed for the reaction. Limiting Reagent = ______ H 2 2) If 3. 9 moles of nitrogen reacts with 12. 1 moles of hydrogen, what is the limiting reagent? 3. 9 mol N 2 x 3 mol H 2 1 mol N 2 = 11. 7 moles of H 2 are Were you given enough H 2? ______ Yes ! (12. 1>11. 7) needed for the reaction. Limiting Reagent = ______ N 2

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.

Excess Reagent (or Excess Reactant) Practice Problem: Find the number of moles of excess reagent from the previous practice problems. 6. 3 mol H 2 x 1 mol N 2 = 2. 1 moles of N 2 are needed for 3 mol H 2 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

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