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CHAPTER 11

CHAPTER 11. Stoichiometry. 11.2 Percent Yield and Concentration. In theory , all 100 kernels should have popped. Did you do something wrong?. +. 100 kernels. 82 popped. 18 unpopped. In theory , all 100 kernels should have popped. Did you do something wrong?. No.

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CHAPTER 11

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  1. CHAPTER 11 Stoichiometry 11.2 Percent Yield and Concentration

  2. In theory, all 100 kernels should have popped. Did you do something wrong? + 100 kernels 82 popped 18 unpopped

  3. In theory, all 100 kernels should have popped. Did you do something wrong? No In real life (and in the lab) things are often not perfect + 100 kernels 82 popped 18 unpopped

  4. Percent yield What you get to eat! + 100 kernels 82 popped 18 unpopped

  5. Percent yield What you get to eat! + 100 kernels 82 popped 18 unpopped

  6. actual yield: the amount obtained in the lab in an actual experiment. theoretical yield: the expected amount produced if everything reacted completely.

  7. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) Heating

  8. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?)

  9. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?) - There is usually some human error, like not measuring exact amounts carefully

  10. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?) - There is usually some human error, like not measuring exact amounts carefully - Maybe the heating time was not long enough; not all the Na2HCO3 reacted

  11. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?) - There is usually some human error, like not measuring exact amounts carefully - Maybe the heating time was not long enough; not all the Na2HCO3 reacted - Maybe Na2CO3 was not completely dry; some H2O(l) was measured too

  12. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?) - There is usually some human error, like not measuring exact amounts carefully - Maybe the heating time was not long enough; not all the Na2HCO3 reacted - Maybe Na2CO3 was not completely dry; some H2O(l) was measured too - CO2 is a gas and does not get measured

  13. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Let’s calculate the percent yield obtained in experiment calculated

  14. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Let’s calculate the percent yield calculated

  15. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Let’s calculate the percent yield calculated

  16. 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) 10.00 g Use the mass of reactant NaHCO3(s) to calculate the mass of the product Na2CO3(s). This is a gram-to-gram conversion:

  17. 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 10.00 g

  18. 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 10.00 g

  19. 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 10.00 g 0.1190 moles

  20. 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 0.1190 moles

  21. 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 0.1190 moles 0.05950 moles

  22. 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 0.05950 moles

  23. 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 0.05950 moles 6.306 g

  24. 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g 10.00 g 6.306 g 0.1190 moles 0.05950 moles For 10.00 g of starting material (NaHCO3), the theoretical yield for Na2CO3 is 6.306 g. The actual yield (measured) is 4.87 g.

  25. 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g For 10.00 g of starting material (NaHCO3), the theoretical yield for Na2CO3 is 6.306 g. The actual yield (measured) is 4.87 g.

  26. Stoichiometry with solutions Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g Convert to moles Reaction of solid zinc with hydrochloric acid: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) Reactions in solution 50.0 mL of a 3.0 M solution Convert to moles

  27. A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess.

  28. A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess. Asked: grams of H2 produced Given: 50.0 mL of 3.0 M HCl reacting with excess zinc Relationships: M = mole/L Mole ratio: 2 moles HCl ~ 1 mole H2 Molar mass of H2 = 1.0079 x 2 = 2.02 g/mole

  29. A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess. Asked: grams of H2 produced Given: 50.0 mL of 3.0 M HCl reacting with excess zinc Relationships: M = mole/L Mole ratio: 2 moles HCl ~ 1 mole H2 Molar mass of H2 = 1.0079 x 2 = 2.02 g/mole Solve:

  30. A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess. Asked: grams of H2 produced Given: 50.0 mL of 3.0 M HCl reacting with excess zinc Relationships: M = mole/L Mole ratio: 2 moles HCl ~ 1 mole H2 Molar mass of H2 = 1.0079 x 2 = 2.02 g/mole Solve:

  31. A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess. Asked: grams of H2 produced Given: 50.0 mL of 3.0 M HCl reacting with excess zinc Relationships: M = mole/L Mole ratio: 2 moles HCl ~ 1 mole H2 Molar mass of H2 = 1.0079 x 2 = 2.02 g/mole Solve:

  32. A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess. Asked: grams of H2 produced Given: 50.0 mL of 3.0 M HCl reacting with excess zinc Relationships: M = mole/L Mole ratio: 2 moles HCl ~ 1 mole H2 Molar mass of H2 = 1.0079 x 2 = 2.02 g/mole Solve: Answer:0.15 grams of H2 are produced

  33. Reaction of solid zinc with hydrochloric acid: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) 50.0 mL of a 3.0 M solution Convert molarity to moles Sometimes the concentration is written in mass percent Vinegar is 5% acetic acid by mass

  34. Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.)

  35. Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.) Asked: grams of acetic acid in 120 mL of vinegar Given: 120 mL of vinegar and 5% acetic acid by mass Relationships:120 mL = 120 g, given a density of 1.0 g/mL

  36. Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.) Asked: grams of acetic acid in 120 mL of vinegar Given: 120 mL of vinegar and 5% acetic acid by mass Relationships: 120 mL = 120 g, given a density of 1.0 g/mL Solve:

  37. Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.) Asked: grams of acetic acid in 120 mL of vinegar Given: 120 mL of vinegar and 5% acetic acid by mass Relationships: 120 mL = 120 g, given a density of 1.0 g/mL Solve:

  38. Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.) Asked: grams of acetic acid in 120 mL of vinegar Given: 120 mL of vinegar and 5% acetic acid by mass Relationships: 120 mL = 120 g, given a density of 1.0 g/mL Solve: Answer:6.0 g of acetic acid.

  39. Obtained from the experiment Calculate using molar masses and mole ratios

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