1 / 19

Lecture No. 5 Limiting Reactant, Theoretical yield and Percentage yield

Lecture No. 5 Limiting Reactant, Theoretical yield and Percentage yield. Lecturer: Amal Abu- Mostafa. Limiting Reactant:. Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly Limiting Reactant Bread Excess Reactants peanut butter and jelly.

emi-perez
Download Presentation

Lecture No. 5 Limiting Reactant, Theoretical yield and Percentage yield

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Lecture No. 5 Limiting Reactant, Theoretical yield and Percentage yield Lecturer: Amal Abu- Mostafa

  2. Limiting Reactant: • Available Ingredients • 4 slices of bread • 1 jar of peanut butter • 1/2 jar of jelly • Limiting Reactant • Bread • Excess Reactants • peanut butter and jelly

  3. Limiting Reactant: • The limiting reactant (or limiting reagent): is the reactant that is entirely consumed (used up)when a reaction goes to completion. • Excess reactant is: a reactant that is not completely consumed in the reaction. • Once one of the reactants is used up, the reaction stops. • This means that: the moles of product are always determined by the starting moles of limiting reactant.

  4. Example 1: • Zinc metal reacts with hydrochloric acid by the following reaction: • Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) • If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, which reactant is the limiting reactant? And how many moles of H2 are produced?

  5. Problem Strategy: • Step 1: Which is the limiting reactant? • The reactant that gives the smaller amount of product is the limiting reactant. • Step 2: You obtain the amount of product actually obtained from the amount of limiting reactant.

  6. Solution for example 1: • Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) • Step 1: From the equation: First: • 1 mol of Zn 1 mol of H2 • 0.30 mol of Zn  ?? Mol of H2 • Moles of H2=0.3× 1 = 0.3 mol 1 • Second: • 2 mol of HCl 1 mol of H2 • 0.52 mol of HCl ?? of H2 • Moles of H2= 0.52 × 1= 0.26 mol 2

  7. Solution for example 1: • So hydrochloric acid HCl must be the limiting reactant because it gave the smaller amount of the product which is H2. • (Zinc is the excess reactant.) • Step 2: Since HCl is the limiting reactant, the amount of H2 produced = 0.26 mol.

  8. Theoretical yield: • The theoretical yield of product: • Is the maximum amount of product that can be obtained by a reaction from given amounts of reactants (in the balanced equation).

  9. Actual yield • Is the amount of products actually obtained when the reaction takes place. • The actual yield of a product may be much less than the theoretical yield for several possible reasons. • It is important to realize that the actual yield is an experimentally determined quantity.

  10. Percentage yield: • The percentage yield of product is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated). measured in lab • Percent yield = Actual yield (g) x 100 Theoretical yield (g) calculated on paper

  11. Example 1: • When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and percentage % yields of KCl. • K2CO3 + 2HCl  2KCl + H2O + CO2 • 45.8 g ?? g • Solution: • The actual yield of KCl = 46.3 g • First we should calculate: • Moles of K2CO3 = m = 45.8 = 0.33 mol • M 138.2

  12. From the equation • K2CO3 + 2HCl  2KCl + H2O + CO2 • 1mol of K2CO3  2 mol of KCl • 0.33 mol of K2CO3  ?? mol of KCl • Moles of KCl = 0.33 × 2 = 0.66 mol • 1 • Theoretical yield of KCl (m) = n × M • = 0.66 × 74.55 • = 49.2 g • Actual yield of KCl = 46.3 g • Percent yield = Actual yield (g) x 100 Theoretical yield (g) • = 46.3 × 100 = 94.1% 49.2

  13. Example 2: • In the reaction: 4 Al + 3 O2→ 2 Al2O3 • A) calculate the theoretical yield of Al2O3 if 54 g of Al is reacted with enough O2 • B) if the experimental yield was 51 g calculate the percentage yield

  14. Solution: • A)Moles of Al = m = 54 = 2 mol M 27 • From the equation: • 4 Al + 3 O2→ 2 Al2O3 • 4 mol of Al → 2 mol of Al2O3 • 2 mol of Al → ?? mol of Al2O3 • Moles of Al2O3 = 2×2= 1 mol 4 • Mw.t of Al2O3= 102 g/mol • Theoretical yield of Al2O3(grams of Al2O3)= n × Mw.t • = 1× 102 = 102 g

  15. B) Percent yield = Actual yield (g) x 100 Theoretical yield (g) • = 51 g × 100 = 50 % 102 g

  16. Learning and check: Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C(g) + O2(g) → 2CO(g) What is the percent yield if 40.0 g CO are produced when 30.0 g of O2 are used? 1) 25.0% 2) 75.0% 3) 76.2%

  17. Solution: • Actual yield of CO = 40 g • Moles of O2 = 30 g = 0.94 mol 32 g/mol • From the equation: • 2C(g) + O2(g) → 2CO(g) • 1 mol of O2 → 2 mol of CO • 0.94 mol of O2 → ?? Mol of CO • Moles of CO = 0.94× 2 = 1.875 mol 1

  18. Solution: • Theoretical yield of CO (m) = n × M w.t = 1.875 × (12+16) = 52.5 g • Percent yield = 40 × 100 = 76.2% 52.5 So the correct answer is (3) 76.2%

  19. Thank you

More Related