NOTES: 12.3 – Limiting Reagent & Percent Yield

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Presentation transcript:

NOTES: 12.3 – Limiting Reagent & Percent Yield

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

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? CS2 + O2  CO2 + SO2 Start by balancing the equation…

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? CS2 + 3O2  CO2 + 2SO2 Now solve the problem…

Which reactant is LIMITING? ● 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? CS2 + 3O2  CO2 + 2SO2 Which reactant is LIMITING?

O2 limits the amount of SO2 that can be produced. CS2 is in excess. ● 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? CS2 + 3O2  CO2 + 2SO2 O2 limits the amount of SO2 that can be produced. CS2 is in excess.

134 g of SO2 can be produced in this reaction. ● 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? CS2 + 3O2  CO2 + 2SO2 134 g of SO2 can be produced in this reaction.

Limiting Reagent Example 3: ● Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2? What mass of the excess reagent will be left over? CH4 + O2  CO2 + H2O Start by balancing the equation…

● Example 3: What mass of CO2 could be formed by the reaction of 8 ● Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2? What mass of the excess reagent will be left over? CH4 + 2O2  CO2 + 2H2O Now determine which reactant is the LIMITING REAGENT…

● Example 3: What mass of CO2 could be formed by the reaction of 8 ● Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2? What mass of the excess reagent will be left over? CH4 + 2O2  CO2 + 2H2O (needed) Which reactant is LIMITING? 8.0 g CH4 requires 32 g of O2… How many grams of oxygen are available…?

● Example 3: What mass of CO2 could be formed by the reaction of 8 ● Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2? What mass of the excess reagent will be left over? CH4 + 2O2  CO2 + 2H2O (needed) Which reactant is LIMITING? 48 g of O2 are available…so O2 is in EXCESS!! 8.0 g CH4 is the LIMITING REAGENT!

● Example 3: What mass of CO2 could be formed by the reaction of 8 ● Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2? What mass of the excess reagent will be left over? CH4 + 2O2  CO2 + 2H2O Use the 8.0 g of CH4 to determine mass of CO2 produced… 22 g CO2 are produced!!

● Example 3: What mass of CO2 could be formed by the reaction of 8 ● Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2? What mass of the excess reagent will be left over? We started with 48 g of O2…we calculated that 32 g of O2 would react…SO, the amount left over is: 48 g – 32 g = 16 g O2

Percent Yield!

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.

● 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.

● Look back at Example 2. We found that 134 g of SO2 could be formed from the reactants. ● Let’s say that in an experiment, you collected 130 g of SO2. What is your percent yield?

Percent Yield Example #1: Calcium carbonate is decomposed by heating. CaCO3 (s)  CaO (s) + CO2 (g) A) What is the theoretical yield of CaO when 24.8 g of CaCO3 decompose? B) When this is done in the lab, 13.1 g of CaO is produced. What is the percent yield of this reaction?

Percent Yield Example #1: CaCO3 (s)  CaO (s) + CO2 (g) A) What is the theoretical yield of CaO when 24.8 g of CaCO3 decompose?

Percent Yield Example #1: CaCO3 (s)  CaO (s) + CO2 (g) A) What is the theoretical yield of CaO when 24.8 g of CaCO3 decompose? = 13.9 g CaO

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?

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%

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) + 3CuSO4 (aq)  Al2(SO4)3 aq) + 3Cu(s)

Percent Yield Example #2: 1st find the “theoretical yield”: 2Al(s) + 3CuSO4 (aq)  Al2(SO4)3 aq) + 3Cu(s)

Percent Yield Example #2: 1st find the “theoretical yield”: 2Al(s) + 3CuSO4 (aq)  Al2(SO4)3 aq) + 3Cu(s) = 6.60 g Cu

Percent Yield Example #2: Now find % yield:

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

CH3COOH + C2H5OH  CH3COOC2H5 + H2O Example #3: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What percent yield of ethyl acetate is this? CH3COOH + C2H5OH  CH3COOC2H5 + H2O

Example #3: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What percent yield of ethyl acetate is this?