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Chemistry 11 Chapter 6

Chemistry 11 Chapter 6. STOICHIOMETRY OF EXCESS QUANTITIES. Introduction: So far. we have assumed that a given reactant is completely used up during the reaction. In reality.

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Chemistry 11 Chapter 6

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  1. Chemistry 11Chapter 6 STOICHIOMETRY OF EXCESS QUANTITIES

  2. Introduction: So far ... • we have assumed that a given reactant is completely used up during the reaction

  3. In reality... • reactions are often carried out in such a way that one or more of the second reactants actually are present in EXCESS amounts.

  4. Definitions • EXCESS REACTANT = the reactant in excess • LIMITING REACTANT = the reactant that completely reacts • THE LIMITING REACTANT determines the yield of the product (how much product(s) will form)

  5. A Simple Analogy • Imagine you work at McDonalds™ … • You have 10 hamburger buns and 5 beef patties • How many regular hamburgers can you make?

  6. Da Answer ... • Indeed, you would get 5 regular hamburgers! • And what was left over?

  7. Connecting the Lingo … • There would be 5 hamburger buns in EXCESS! • Therefore, the beef patty is known as the LIMITING ingredient since it “limits” or determines how many regular buns can be made!

  8. Note that • we do not predict based on the number of hamburger buns

  9. Example 1 • If 20.0 g of hydrogen gas react with 100.0 g of oxygen, which reactant is present in excess and by how many grams?

  10. Step 1 … • The balanced equation:

  11. Step 2 … • First PREDICT which reactant is limiting (it’s ok if you predict wrong) • USUALLY the reactant with the least number of moles is limiting (but not always)

  12. Convert masses to moles Number of moles of H2 present = 20.0 g x 1 mol H2 = 10 mol H2 2.0 g H2 Number of moles of O2 present = 100.0 g x 1 mol O2 = 3.125mol O232.0 g O2 Let’s make a prediction ...

  13. Prediction: O2 is limiting Mass of H2 that reacts with 100.0 g O2 = 100.0 g O2 x 1 mol O2 x 2 mol H2 x 2.0 g H2 32.0 g O21 mol O2 1 mol H2 = 12.5 g H2

  14. Analysing the numbers … • What we have: 100.0 g O2 and 20.0 g H2 • We predict O2 is limiting (i.e. all 100.0 g reacted) • We calculated that we would need 12.5 g H2 • Is the prediction correct ?

  15. Da Answer (again!) • Yes!! Prediction is correct • only 12.5 g H2 is required, so we have an excess of 7.5 g H2 (20.0 g - 12.5 g) • so H2 is in EXCESS of 7.5 g.

  16. The other side of the coin So what if we predicted that H2 was limiting? Mass of O2 that reacts with 20.0 g H2 = 20.0 g H2 x 1 mol H2 x 1 mol O2x 32.0 g O2 2.0 g H2 2 mol H2 1 mol O2 = 160.0 g O2

  17. Therefore … • If ALL 20.0 g of H2 were to completely react we would need 160.0 g of O2 • BUT we only have 100.0 g of O2 • So the prediction that H2 limiting is INCORRECT!

  18. Example 2. If 79.1 g of Zn reacts with 1.05 L of 2.00 M HCl, a) Which reactant is in excess and by how much? b) What is the mass of each product?

  19. a) which reactant is excess? The balanced equation: Zn + 2HCl  ZnCl2 + H2 79.1 g 1.05 L, 2.00 M x g y g   1.21 mol 2.10 mol (what we HAVE)

  20. Prediction: Zn is limiting Moles of HCl required = 1.21 mol Zn x 2 mol HCl = 2.43 mol HCl 1 mol Zn Therefore 2.42 mol HCl would be required to react with 1.21 mol Zn. We ONLY have 2.10 mol HCl So is our prediction correct?

  21. Uh Oh! You’re wrong! • We would need more HCl (2.42 mol) than what we have (2.10 mol) if all the Zn were to react • Thus: Zn is in excess, and HCl is limiting!

  22. To find how much in excess: We must find how many moles of Zn is required to react with 2.10 molHCl Mol of Zn = 2.10 mol HCl x 1 molZn 2 molHCl = 1.05 mol Zn Excess Zn = 1.21 - 1.05 = 0.16 mol Zn

  23. b) mass of products? Since HCl is limiting we MUST use this amount to calculate the mass of products x g ZnCl2 = 2.10 mol HCl x 1 mol ZnCl2 x 136.4 g 2 mol HCl 1 mol ZnCl2 = 143 g ZnCl2 y g H2 = 2.10 mol HCl x 1 mol H2 x 2.0 g 2 mol HCl 1 mol H2 = 2.1 g H2

  24. Example 3: • 3.00 L of 0.1 M NaCl reacts with 2.50 L of 0.125 M AgNO3. Calculate the yield of solid AgCl (in grams) that will be produced. • This problem requires us to determine how much product (AgCl) will form, so we will need to first determine which reactant is limiting.

  25. The balanced equation: NaCl(aq) + AgNO3(aq) NaNO3(aq)+ AgCl(s) 3.00 L 2.50 L ? g 0.1M 0.125 M   0.300 mol 0.325 mol Limiting Excess (since 1:1 ratio)

  26. NaCl limiting ... Therefore: mol NaCl = mol AgCl = 0.300 mol (also 1:1 ratio) Mass of AgCl = 0.300 mol AgCl x 143.5 g AgCl 1 mol AgCl = 43.1 g AgCl

  27. Percent Yield • Often 100% of the expected amount of product cannot be obtained from a reaction • The term “Percent Yield” is used to describe the amount of product actually obtained as a percentage of the expected amount

  28. Reasons for reduced yields A) the reactants may not all react because: i) not all of the pure material actually reacts ii) the reactants may be impure B) Some of the products are lost during procedures such as solvent extraction, filtration etc

  29. The equation: Percent Yield = ACTUAL YIELD x 100% THEORETICAL YIELD • Actual yield = amount of product obtained (determined experimentally) • Theoretical yield = amount of product expected (determined from calculations based on the stoichiometry of the reaction) • The amounts may be expressed in g, mol, molecules

  30. Types of calculations A) Find the percentage yield, given the mass of reactant used and mass of product formed B) Find the mass of product formed, given the mass of reactant used and the percentage yield C) Find the mass of reactant used, given the mass of product formed and percentage yield • Note that the percentage yield must be less than 100% • But when calculating the theoretical yield assume a 100% yield

  31. Example 1 When 15.0 g of CH4 is reacted with an excess of Cl2 according to the reaction: CH4 + Cl2 CH3Cl + HCl a total of 29.7 g of CH3Cl is formed. Calculate the percentage yield.

  32. The solution ... The actual yield of CH3Cl = 29.7 g To find the theoretical yield of CH3Cl: (assuming a 100% yield) g of CH3Cl = 15.0 g CH4 x 1 mol CH4x 1 mol CH3Cl x 50.5 g 16.0g CH4 1 mol CH4 1 mol CH3Cl = 47.34 g

  33. Then: Percentage yield = actual yield x 100% theoretical yield = 29.7 g x 100% = 62.7 % 47.34 g

  34. Example deux! What mass of K2CO3 is produced when 1.50 g of KO2 is reacted with an excess of CO2 if the reaction has a 76.0% yield? The reaction is: 4KO2(s) + 2 CO2(g) 2K2CO3(s) + 3O2(g)

  35. The solution: We are looking for the actual yield (some idiot forgot to weigh and record the mass of product!) First calculate the mass of K2CO3 produced (assuming a 100% yield) i.e. the theoretical yield g of K2CO3 = 1.50 g KO2 x 1 mol KO2 x 2 mol K2CO3 x 138.2 g 71.1 g KO2 4 mol KO2 1 mol K2CO3 = 1.458 g actual yield = 76.0 % x 1.458 g = 0.760 x 1.458 = 1.11 g K2CO3

  36. Last (but not least) example ... What mass of CuO is required to make 10.0 g of Cu according to the reaction 2NH3 + 3CuO  N2 + 3Cu+ 3H2O if the reaction has 58.0 % yield?

  37. Here we go again … Actual yield = 10.0 g Cu From the percentage yield equation, calculate the theoretical yield of Cu. Theoretical yield of Cu = Actual yield x 100% Percentage yield = 10.0 g x 100 % 58.0 % = 17.24 g

  38. Now find the mass of CuO: Use this theoretical yield and find the mass of CuO that would be needed: g CuO = 17.24 g Cu x 1 mol Cu x 3 mol CuO x 79.5g 63.5 g Cu 3 mol Cu 1 mol CuO = 21.6 g CuO

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