Lesson 22: Theoretical Yield Actual Yield Percent Yield

Transcription

1 Lesson 22: Theoretical Yield Actual Yield Percent Yield

2 Do Now (5pts) Copy down info from CJ board. Answer questions in Box of Lesson 22 note packet. You have a test in one week it ll be multiple choice and cumulative. Quiz corrections will be due by Thursday March 29.

3 2 4 2 flags + pole + 4 wheels + car body car If you had 62 flags, 37 poles, 02 wheels, and 30 car bodies, how many complete toy cars could you make? What reagent is limiting? 62 flags / 2 per car = 3 37 poles / per car = wheels / 4 per car = car bodies / per car = 30 Has the smallest equivalent, the wheels are the limiting reagent You can only make 25 complete cars

4 B F F B F 4 If you started with 30 B beads, 60 F beads, 00 purple beads, 90 yellow beads, 24 pink beads, 50 blue beads, and 45 empty bracelets, how many complete BFF bracelets could you make? What reagent is limiting? 30 B beads / per bracelet = F beads / 2 per bracelet = purple beads / 4 per bracelet = yellow beads / 4 per bracelet = pink beads / per bracelet = empty bracelets / per bracelet = 45 Has the smallest equivalent, the yellow beads are the limiting reagent. You can only make 22 bracelets.

5 SiO 2 (s) + 4HF (g) SiF 4 (g) + 2H 2 O (l) If you started with 0.90 mol of silicon dioxide and 3.5 mol of HF, what is the limiting reagent? 0.90 mol silicon dioxide / = mol HF / 4 = Has the smallest equivalent, HF is the limiting reagent. What is the theoretical yield of water (in moles)? 3.5 mol HF 2 mol H 2 O = 4 mol HF molar ratio.76 mol H 2 O

6 Review: Limiting and Excess Reagents One reactant almost always limits the amount of product produced in a reaction. Once one of the reactant is used up,. then no more product can be formed The substance that is used up first is the. limiting reagent Identifying the limiting reagent: calculate the. molar equivalents The limiting reagent has the lesser equivalent (you will run out of that reactant first).

7 SiO 2 + 4HF 2 SiF H 2 O 4.5 mol 6.0 mol What is the limiting reagent? HF Amount that you start with Co-efficient of the reactant SiO vs. HF This number is SMALLER, therefore it is the limiting reagent These numbers are called molar equivalents

8 Finding the molar equivalent of each reactant: Amount that you start with This has to be in compounds or moles. Co-efficient of the reactant

9 N 2 H 4 (l) + 2 H 2 O 2 (l) N 2 (g) + 4 H 2 O (g) mol 5.2 mol What is the limiting reagent? N 2 H 4 N 2 H 4 H 2 O 2 Amount that you start with Co-efficient of the reactant vs This number is SMALLER, therefore it is the limiting reagent These numbers are called molar equivalents

10 Let s look at #3 on p.3

11 Methanol, CH 3 OH is the simplest of alcohols. It is synthesized by the reaction of hydrogen and carbon monoxide. CO (g) + 2H 2 (g) CH 3 OH If 500 mol CO and 750 mol H 2 are present, which is the limiting reagent? 500 mol CO 750 mol H H 2 is the limiting reagent

12 Methanol, CH 3 OH is the simplest of alcohols. It is synthesized by the reaction of hydrogen and carbon monoxide. CO (g) + 2H 2 (g) CH 3 OH How many moles of the excess reagent remain unchanged? Start with the limiting reagent: 750 mol H 2 mol CO 2 mol H2 = 375 mol CO will be used up Amount we started with amount consumed in reaction 500 mol CO 375 mol CO = 25 mol CO

13 Methanol, CH 3 OH is the simplest of alcohols. It is synthesized by the reaction of hydrogen and carbon monoxide. CO (g) + 2H 2 (g) CH 3 OH How many moles of CH 3 OH are formed? Start with the limiting reagent: 750 mol H 2 mol CH 3 OH 2 mol H2 = 375 mol CH 3 OH will be produced

14 Let s look at the second problem on p.3

15 The black oxide of iron, Fe 3 O 4, occurs in nature as the mineral magnetite. The substance can also be used in the laboratory by the reaction between red-hot iron and steam according to the following reaction: LR 3Fe(s) + 4H 2 O (g) Fe 3 O 4 (s) + 4H 2 (g) When 36 g of H 2 O are mixed with 67.0 g Fe, which is the limiting reagent? FIRST find how many moles of each reactant you have. THEN calculate the mole equivalents. 36 g H 2 O molar mass mol H 2 O 8 g H 2 O = 2 mol H 2 O mole equivalent 2 mol H 2 O 4 = g Fe molar mass mol Fe g Fe =.9964 mol Fe mole equivalent.9964 mol Fe 3 =

16 LR 3Fe(s) + 4H 2 O (g) Fe 3 O 4 (s) + 4H 2 (g) 67.0 g ER 36 g What mass in grams of black iron oxide is produced? This is a grams grams conversion. Start with the limiting reagent: 67.0 g Fe mol Fe g Fe molar mass mol Fe 3 O 4 mol Fe g Fe 3 O 4 = 3 Molar ratio of reactant to product in balanced reaction mol Fe 3 O 4 molar mass g Fe 3 O 4

17 LR 3Fe(s) + 4H 2 O (g) Fe 3 O 4 (s) + 4H 2 (g) 67.0 g ER 36 g What mass in grams of excess reagent remains when the reaction is completed? Start with the limiting reagent: 67 g Fe mol Fe g Fe mol H 2 O mol Fe Amount we started with amount consumed in reaction 8 g H 2 O = mol H 2 O 29 g H 2 O consumed during the reaction 36g water to start 29 g used up = 7 g water left over

18 Do Now (5pts) Copy down info from CJ board. Take out your HW for me to check. Answer questions in Box 2 of Lesson 22 note packet (last page). You have a test in one week it ll be multiple choice and cumulative. Quiz corrections will be due by Thursday March 29.

19 THEORETICAL YIELD Theoretical yield is the maximum amount of product that can be formed when the limiting reagent is completely consumed. depends solely on the amount of the limiting reagent Theoretical yield and is calculated using stoichiometry. beginning from the limiting reagent.

20 THEORETICAL YIELD mole ratio Just as the allows us to calculate limiting and excess reagents and predict masses, we can also the mass of products in any given reaction. the maximum quantity of product possible if the reaction proceeds to completion. The theoretical yield represents

21 ACTUAL YIELD measured amount of a product The obtained from a reaction is called the actual yield (or experimental yield) of that product.

22 PERCENT YIELD Comparing the theoretical and actual yield helps chemists determine the reaction s. efficiency The percent yield represents the of ratio the actual yield to the theoretical yield. Percentage yield = actual yield theoretical yield 00

23 Let s look at the # on p.5

24 CH 4 + 2O 2 CO 2 + 2H 2 O How many grams of H 2 O are expected when 8.00 grams of CH 4 reacts with an excess of O 2? Start with the limiting reagent: 8.00 g CH 4 mol CH g CH 4 2 mol H 2 O mol CH 4 g H 2 O = 8.06 mol H 2 O 8.0 g H 2 O Theoretical Yield

25 CH 4 + 2O 2 CO 2 + 2H 2 O If only 7.0 grams of H 2 O are produced, what is the percent yield of this reaction? 7.0 g H 2 O ACTUAL Yield 8.0 g H 2 O Theoretical Yield 7.0 / 8.0 = % yield Percentage yield = actual yield theoretical yield 00

26 Let s look at the #2 on p.5

27 Starting amounts: C 6 H 6 (l) + Cl 2 (g) C 6 H 5 Cl (l) + HCl (g) 52.6 g 60.2 g Which is the limiting reagent? WHAT DO I DO? First, convert from grams to moles using molar mass. Then, calculate mole equivalents.

28 Starting amounts: C 6 H 6 (l) + Cl 2 (g) C 6 H 5 Cl (l) + HCl (g) 52.6 g 60.2 g LR ER Which is the limiting reagent? 52.6 g C 6 H 6 mol C 6 H 6 = mol C g C 6 H 6 H / = g Cl 2 mol Cl 2 = mol Cl g Cl / = 0.849

29 Starting amounts: C 6 H 6 (l) + Cl 2 (g) C 6 H 5 Cl (l) + HCl (g) 52.6 g 60.2 g LR ER Determine the amount of excess reagent reacted and amount remaining. Start with the limiting reagent: 52.6 g C 6 H 6 mol C 6 H 6 mol Cl g Cl g C 6 H 6 mol C 6 H 6 mol Cl g to start 47.7 g consumed = 2.5 g of Cl 2 in excess 47.7 g chlorine consumed in reaction

30 Starting amounts: C 6 H 6 (l) + Cl 2 (g) C 6 H 5 Cl (l) + HCl (g) 52.6 g 60.2 g LR ER Calculate the theoretical yield of product chlorobenzene. Start with the limiting reagent: 52.6 g C 6 H g C 6 H 6 mol C 6 H 5 Cl mol C 6 H 6 mol C 6 H g C 6 H 5 Cl mol C 6 H 5 Cl Theoretically made if the reaction was 00% efficient = g C 6 H 5 Cl BUT REACTIONS ARE NOT ALWAYS 00% EFFICIENT!

31 C 6 H 6 (l) + Cl 2 (g) C 6 H 5 Cl (l) + HCl (g) If 73.9 g of chlorobenzene (C 6 H 5 Cl) was actually produced, what was the percent yield? Actual yield = 73.9 g C 6 H 5 Cl Theoretical yield = g C 6 H 5 Cl 73.9 / = % yield Percentage yield = actual yield theoretical yield 00