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NOTES: 12.3 – Limiting Reagent & Percent Yield. Calculations need to be based on the limiting reactant. ● Example 1: Suppose a box contains 87 bolts,

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Presentation on theme: "NOTES: 12.3 – Limiting Reagent & Percent Yield. Calculations need to be based on the limiting reactant. ● Example 1: Suppose a box contains 87 bolts,"— Presentation transcript:

1 NOTES: 12.3 – Limiting Reagent & Percent Yield

2 Calculations need to be based on the limiting reactant. ● Example 1: Suppose a box contains 87 bolts, 110 washers and 99 nails. How many sets of 1 bolt, 2 washers and 1 nail can you use to create? What is the limiting factor? 55 sets; washers limit the amount

3 Calculations need to be based on the limiting reactant. ● Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen? CS 2 + O 2  CO 2 + SO 2 Start by balancing the equation…

4 Calculations need to be based on the limiting reactant. ● Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen? CS 2 + 3O 2  CO 2 + 2SO 2 Now solve the problem…

5 ● Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen? CS 2 + 3O 2  CO 2 + 2SO 2 Which reactant is LIMITING?

6 ● Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen? CS 2 + 3O 2  CO 2 + 2SO 2 O 2 limits the amount of SO 2 that can be produced. CS 2 is in excess.

7 ● Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen? CS 2 + 3O 2  CO 2 + 2SO g of SO 2 can be produced in this reaction.

8 ● Example 3: What mass of CO 2 could be formed by the reaction of 8.0 g CH 4 with 48 g O 2 ? What mass of the excess reagent will be left over? CH 4 + O 2  CO 2 + H 2 O Start by balancing the equation… Limiting Reagent Example 3:

9 ● Example 3: What mass of CO 2 could be formed by the reaction of 8.0 g CH 4 with 48 g O 2 ? What mass of the excess reagent will be left over? CH 4 + 2O 2  CO 2 + 2H 2 O Now determine which reactant is the LIMITING REAGENT…

10 ● Example 3: What mass of CO 2 could be formed by the reaction of 8.0 g CH 4 with 48 g O 2 ? What mass of the excess reagent will be left over? CH 4 + 2O 2  CO 2 + 2H 2 O Which reactant is LIMITING? 8.0 g CH 4 requires 32 g of O 2 … How many grams of oxygen are available…? (needed)

11 ● Example 3: What mass of CO 2 could be formed by the reaction of 8.0 g CH 4 with 48 g O 2 ? What mass of the excess reagent will be left over? CH 4 + 2O 2  CO 2 + 2H 2 O Which reactant is LIMITING? 48 g of O 2 are available…so O 2 is in EXCESS!! 8.0 g CH 4 is the LIMITING REAGENT! (needed)

12 ● Example 3: What mass of CO 2 could be formed by the reaction of 8.0 g CH 4 with 48 g O 2 ? What mass of the excess reagent will be left over? CH 4 + 2O 2  CO 2 + 2H 2 O Use the 8.0 g of CH 4 to determine mass of CO 2 produced… 22 g CO 2 are produced!!

13 ● Example 3: What mass of CO 2 could be formed by the reaction of 8.0 g CH 4 with 48 g O 2 ? What mass of the excess reagent will be left over? We started with 48 g of O 2 …we calculated that 32 g of O 2 would react…SO, the amount left over is: 48 g – 32 g = 16 g O 2

14 Percent Yield!

15 Many chemical reactions do not go to completion (reactants are not completely converted to products). Percent Yield: indicates what percentage of a desired product is obtained.

16 ● So far, the masses we have calculated from chemical equations were based on the assumption that each reaction occurred 100%. ● The THEORETICAL YIELD is the amount calculated from the balance equation. ● The ACTUAL YIELD is the amount “actually” obtained in an experiment.

17 ● Look back at Example 2. We found that 134 g of SO 2 could be formed from the reactants. ● Let’s say that in an experiment, you collected 130 g of SO 2. What is your percent yield?

18 Percent Yield Example #1: Calcium carbonate is decomposed by heating. CaCO 3 (s)  CaO (s) + CO 2 (g) A) What is the theoretical yield of CaO when 24.8 g of CaCO 3 decompose? B) When this is done in the lab, 13.1 g of CaO is produced. What is the percent yield of this reaction?

19 Percent Yield Example #1: CaCO 3 (s)  CaO (s) + CO 2 (g) A) What is the theoretical yield of CaO when 24.8 g of CaCO 3 decompose?

20 Percent Yield Example #1: CaCO 3 (s)  CaO (s) + CO 2 (g) A) What is the theoretical yield of CaO when 24.8 g of CaCO 3 decompose? = 13.9 g CaO

21 Percent Yield Example #1: B) When this is done in the lab, 13.1 g of CaO is produced. What is the percent yield of this reaction?

22 Percent Yield Example #1: B) When this is done in the lab, 13.1 g of CaO is produced. What is the percent yield of this reaction? = 94.2%

23 Percent Yield Example #2: What is the percent yield if 4.65 g of copper is produced when 1.87 g of aluminum reacts with an excess of copper(II) sulfate? 2Al (s) + 3CuSO 4 (aq)  Al 2 (SO 4 ) 3 aq) + 3Cu (s)

24 Percent Yield Example #2: 1 st find the “theoretical yield”: 2Al (s) + 3CuSO 4 (aq)  Al 2 (SO 4 ) 3 aq) + 3Cu (s)

25 Percent Yield Example #2: 1 st find the “theoretical yield”: 2Al (s) + 3CuSO 4 (aq)  Al 2 (SO 4 ) 3 aq) + 3Cu (s) = 6.60 g Cu

26 Percent Yield Example #2: Now find % yield:

27 Percent Yield Example #2: Now find % yield: = 70.5%

28 Example #3: A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5. What percent yield of ethyl acetate is this? CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5 + H 2 O

29 Example #3: A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5. What percent yield of ethyl acetate is this?


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