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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.

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

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  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

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