html5-img
1 / 84

Stoker Exercises

How is stoichiometry like baking cookies?. Stoker Exercises. A recipe indicates the amount of each ingredient and the procedure used to produce a certain number of cookies. By looking at the list of ingredients, you can predict what the finished cookie will be like. ?. Stoichiometry.

starbuck
Download Presentation

Stoker Exercises

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. How is stoichiometry like baking cookies? Stoker Exercises A recipe indicates the amount of each ingredient and the procedure used to produce a certain number of cookies. By looking at the list of ingredients, you can predict what the finished cookie will be like. ?

  2. Stoichiometry 10.45 How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 1.772 g of the first-listed reactant? a) SiO2+ 3C 2CO + SiC g C = ? 1.772 g SiO2 1 mol SiO2 3 mol C 12.01 g C X ------------------ X ------------------ X ------------------ 60.09 g SiO2 1 mol SiO2 1 mol C = 1.062 g C

  3. Stoichiometry 10.45 How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 1.772 g of the first-listed reactant? b) 5 O2 + C3H8 3CO2 + 4H2O g C3H8= ? 1.772 g O2 1 mol O2 1 mol C3H8 44.09 g C3H8 X ------------------ X ------------------ X ------------------ 32.00 g O2 5 mol O2 1 mol C3H8 = 0.4883 g C3H8

  4. Stoichiometry 10.45 How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 1.772 g of the first-listed reactant? c) CH4 + 4Cl2 4HCl + CCl4 g Cl2= ? 1.772 g CH4 1 mol CH4 4 mol Cl2 70.90 g Cl2 X ------------------ X ------------------ X ------------------ 16.04 g CH4 1 mol CH4 1 mol Cl2 = 31.33 g Cl2

  5. Stoichiometry 10.45 How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 1.772 g of the first-listed reactant? d) 3NO2 + H2O  2HNO3 + NO g H2O= ? 1.772 g NO2 1 mol NO2 1 mol H2O 18.02 g H2O X ------------------ X ------------------ X ------------------ 46.01 g NO2 3 mol NO2 1 mol H2O = 0.2313 g H2O

  6. Stoichiometry 10.46 a How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 12.56 g of the first-listed reactant? a) H2O2 + H2S  2H2O + S g H2S = ? 12.56 g H2O2 1 mol H2O2 1 mol H2S 34.09 g H2S X ------------------ X ------------------ X ------------------ 34.02 g H2O2 1 mol H2O2 1 mol H2S = 12.59 g H2S

  7. Stoichiometry 10.46 b How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 12.56 g of the first-listed reactant? b) 4NH3 + 3O2  2N2 + 6H2O g O2 = ? 12.56 g NH3 1 mol NH3 3 mol O2 32.00 g O2 X ------------------ X ------------------ X ------------------ 17.04 g NH3 4 mol NH3 1 mol O2 = 17.69 g O2

  8. Stoichiometry 10.46 c How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 12.56 g of the first-listed reactant? c) Mg + 2HCl  MgCl2 + H2 g HCl = ? 12.56 g Mg 1 mol Mg 2 mol HCl 36.46 g HCl X ------------------ X ------------------ X ------------------ 24.30 g Mg 1 mol Mg 1 mol HCl = 37.69 g HCl

  9. Stoichiometry 10.46 d How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 12.56 g of the first-listed reactant? d) 6HCl + 2Al  3H2 + 2AlCl3 g Al = ? 12.56 g HCl 1 mol HCl 2 mol Al 26.98 g Al X ------------------ X ------------------ X ------------------ 36.46 g HCl 6 mol HCl 1 mol Al = 3.098 g Al

  10. 10.47 Silicon carbide, SiC, used as an abrasive on sandpaper, is prepared using the following chemical reaction: SiO2(s) + 3C(s)  SiC(s) + 2CO(g) Stoichiometry a) How many grams of SiO2 are needed to react with 1.50 moles of C? a) SiO2 + 3C  SiC + 2CO g SiO2 = ? 1.50 mol C 1 mol SiO2 60.1 g SiO2 = 30.1 g SiO2 X ------------------ X ------------------ 3 mol C 1 mol SiO2

  11. 10.47 Silicon carbide, SiC, used as an abrasive on sandpaper, is prepared using the following chemical reaction: SiO2(s) + 3C(s)  SiC(s) + 2CO(g) Stoichiometry b) How many grams of CO are produced when 1.37 moles of SiO2 react? b) SiO2 + 3C  SiC + 2CO g CO = ? 1.37 mol SiO2 2 mol CO 28.0 g CO = 76.7 g CO X ------------------ X ------------------ 1 mol SiO2 1 mol CO

  12. 10.47 Silicon carbide, SiC, used as an abrasive on sandpaper, is prepared using the following chemical reaction: SiO2(s) + 3C(s)  SiC(s) + 2CO(g) Stoichiometry c) How many grams of SiC are produced at the same time that 3.33 moles of CO are produced? c) SiO2 + 3C  SiC + 2CO g SiC = ? 3.33 mol CO 1 mol SiC 40.1 SiC = 66.8 g SiC X ------------------ X ------------------ 2 mol CO 1 mol SiC

  13. 10.47 Silicon carbide, SiC, used as an abrasive on sandpaper, is prepared using the following chemical reaction: SiO2(s) + 3C(s)  SiC(s) + 2CO(g) Stoichiometry d) How many grams of C must react in order to produce 0.575 mole of SiC? d) SiO2 + 3C  SiC + 2CO g C = ? 0.575 mol SiC 3 mol C 12.0 g C = 20.7 g C X ------------------ X ------------------ 1 mol SiC 1 mol C

  14. 10.48 The inflating gas for automobile air bags is nitrogen (N2), generated from the decomposition of sodium azide (NaN3). The equation for the decomposition reaction is: 2NaN3(s)  2 Na(s) +3N2(g) Stoichiometry a) How many grams of NaN3 must decompose to produce 3.57 moles of N2? a) 2NaN3 2 Na + 3N2 g NaN3 = ? 3.57 mol N2 2 mol NaN3 65.0 NaN3 = 155 g NaN3 X ------------------ X ------------------ 3 mol N2 1 mol NaN3

  15. 10.48 The inflating gas for automobile air bags is nitrogen (N2), generated from the decomposition of sodium azide (NaN3). The equation for the decomposition reaction is: 2NaN3(s)  2 Na(s) + 3N2(g) Stoichiometry b) How many grams of NaN3 must decompose to produce 3.57 moles of Na? b) 2NaN3 2 Na + 3N2 g NaN3 = ? 3.57 mol Na 2 mol NaN3 65.0 NaN3 = 232 g NaN3 X ------------------ X ------------------ 2 mol Na 1 mol NaN3

  16. 10.48 The inflating gas for automobile air bags is nitrogen (N2), generated from the decomposition of sodium azide (NaN3). The equation for the decomposition reaction is: 2NaN3(s)  2 Na(s) + 3N2(g) Stoichiometry c) How many grams of Na are produced at the same time that 5.40 moles of N2 are produced? c) 2NaN3 2 Na  3N2 g Na = ? 5.40 mol N2 2 mol Na 23.0 Na = 82.8 g Na X ------------------ X ------------------ 3 mol N2 1 mol Na

  17. 10.48 The inflating gas for automobile air bags is nitrogen (N2), generated from the decomposition of sodium azide (NaN3). The equation for the decomposition reaction is: 2NaN3(s)  2 Na(s) + 3N2(g) Stoichiometry d) How many moles of NaN3 must decompose in order to produce 10.00 g of N2? d) 2NaN3 2 Na  3N2 mol NaN3 = ? 10.00 g N2 1 mol N2 2 mol NaN3 = 0.2379 mol NaN3 X ------------------ X ------------------ 28.02 g N2 3 mol N2

  18. Stoichiometry 10.49 One way to remove gaseous carbon dioxide (CO2) from the air in a spacecraft is to let canisters of solid lithium hydroxide (LiOH) absorb it according to the reaction 2LiOH + CO2 Li2CO3 + H2O. Based on this equation, calculate how many grams of LiOH must be used to achieve the following: a) absorb 4.50 moles of CO2 2LiOH + CO2 Li2CO3 + H2O g LiOH = ? 4.50 mol CO2 2 mol LiOH 23.9 g LiOH 215 g LiOH X ------------------ X ------------------ 1 mol CO2 1 mol LiOH

  19. Stoichiometry 10.49 One way to remove gaseous carbon dioxide (CO2) from the air in a spacecraft is to let canisters of solid lithium hydroxide (LiOH) absorb it according to the reaction 2LiOH + CO2 Li2CO3 + H2O. Based on this equation, calculate how many grams of LiOH must be used to achieve the following: b) absorb 3.00 X 1024 molec. CO2 2LiOH + CO2 Li2CO3 + H2O g LiOH = ? 3.00 X 1024 molc. CO2 1 mol CO2 23.9 g LiOH 2 mole LiOH X ------------------ X ------------------ X ------------------ 1 mol LiOH 6.02 X 1023 molec. CO2 1 mol CO2 238 g LiOH

  20. Stoichiometry 10.49 One way to remove gaseous carbon dioxide (CO2) from the air in a spacecraft is to let canisters of solid lithium hydroxide (LiOH) absorb it according to the reaction 2LiOH + CO2 Li2CO3 + H2O. Based on this equation, calculate how many grams of LiOH must be used to achieve the following: c) produce 10.0 g H2O 2LiOH + CO2 Li2CO3 + H2O g LiOH = ? 10.0 g H2O 1 mol H2O 2 mole LiOH 23.9 g LiOH X ------------------ X ------------------ X ------------------ 18.0 g H2O 1 mol H2O 1 mol LiOH 26.6 g LiOH

  21. Stoichiometry 10.49 One way to remove gaseous carbon dioxide (CO2) from the air in a spacecraft is to let canisters of solid lithium hydroxide (LiOH) absorb it according to the reaction 2LiOH + CO2 Li2CO3 + H2O. Based on this equation, calculate how many grams of LiOH must be used to achieve the following: d) produce 10.0 g of Li2CO3 2LiOH + CO2 Li2CO3 + H2O g LiOH = ? 10.0 g Li2CO3 1 mol Li2CO3 2 mole LiOH 23.9 g LiOH X ------------------ X ------------------ X ------------------ 73.9 g Li2CO3 1 mol Li2CO3 1 mol LiOH 6.47 g LiOH

  22. Stoichiometry 10.50 a) Tungsten (W) metal, used to make incandescent light bulb filaments, is produced by the reaction WO3(s) + 3H2(g)  W(s) + 3H2O(l). Based on this equation, calculate how many grams of WO3 are needed to produce each of the following: a) 10.00 g of W b) 1.00 X 109 molec. H2O b) 2.53 moles of H2O d) 250 000 atoms of W WO3(s) + 3H2(g)  W(s) + 3H2O(l) g WO3 = ? 10.00 g W 1 mol W 1 mol WO3 231.8 g WO3 X ------------------ X ------------------ X ------------------ 183.8 g W 1 mol W 1 mol WO3 12.61 g WO3

  23. Stoichiometry 10.50 b) Tungsten (W) metal, used to make incandescent light bulb filaments, is produced by the reaction WO3(s) + 3H2(g)  W(s) + 3H2O(l). Based on this equation, calculate how many grams of WO3 are needed to produce each of the following: a) 10.00 g of W b) 1.00 X 109 molec. H2O c) 2.53 moles of H2O d) 250 000 atoms of W WO3(s) + 3H2(g)  W(s) + 3H2O(l) g WO3 = ? 1.00 X 109 molec. H2O 1 mol H2O 1 mol WO3 231.8 g WO3 X ------------------ X ------------------ X ------------------ 6.02 X 1023 molec. H2O 3 mol H2O 1 mol WO3 1.28 X 10-13 g WO3

  24. Stoichiometry 10.50 c) Tungsten (W) metal, used to make incandescent light bulb filaments, is produced by the reaction WO3(s) + 3H2(g)  W(s) + 3H2O(l). Based on this equation, calculate how many grams of WO3 are needed to produce each of the following: a) 10.00 g of W b) 1.00 X 109 molec. H2O c) 2.53 moles of H2O d) 250 000 atoms of W WO3(s) + 3H2(g)  W(s) + 3H2O(l) g WO3 = ? 2.53 mol H2O 1 mol WO3 231.8 gWO3 = 195 g WO3 X ------------------ X ------------------ 3 mol H2O 1 mol WO3

  25. Stoichiometry 10.50 d) Tungsten (W) metal, used to make incandescent light bulb filaments, is produced by the reaction WO3(s) + 3H2(g)  W(s) + 3H2O(l). Based on this equation, calculate how many grams of WO3 are needed to produce each of the following: a) 10.00 g of W b) 1.00 X 109 molec. H2O c) 2.53 moles of H2Od) 250 000 atoms of W WO3(s) + 3H2(g)  W(s) + 3H2O(l) g WO3 = ? 2.5 X 105 atoms W 1 mol W 1 molWO3 231.8 g WO3 X ------------------ X ------------------ X ------------------ 6.0 X 1023 atoms W 1 mol W 1 mol WO3 = 9.7 X 10 -17 g WO3

  26. Stoichiometry

  27. Stoichiometry

  28. Stoichiometry K2S2O3 + 4Cl2 + 5H2O  2 KHSO4 + 8HCl mol K2S2O3 = ? 2.500 g HCl 1 mol HCl 1 molK2S2O3 X ------------------ X ------------------ = 0.008571 mol K2S2O 36.46 g HCl 8 mol HCl

  29. Stoichiometry

  30. Stoichiometry

  31. Stoichiometry

  32. Stoichiometry 10.53 How many grams of sodium are needed to react completely with 16.5 g of sulfur in the synthesis of Na2S? 2Na + S  Na2S g Na = ? 16.5 g S 2 mol Na 23.0 g Na 1 mol S 16.5 g S X ------------- X ------------- X ------------- 1 mol S 1 mol Na 32.1 g S = 23.6 g Na

  33. Stoichiometry 10.54 How many grams of beryllium are needed to react completely with 45.0 g of nitrogen in the synthesis of Be3N2? 3Be + N2 Be3N2 g Be = ? 45.0 g N2 3 mol Be 9.01 g Be 1 mol N2 45.0 g N2 X ------------- X ------------- X ------------- 1 mol N2 1 mol Be 28.0 g N2 = 43.4 g Be

  34. Stoichiometry 10.55 When chromium metal reacts with chlorine gas, a violet solid with the formula CrCl3 is formed: 2Cr + 3Cl2 2CrCl3. How many grams of Cr and how many grams of Cl2 are needed to produce 200.0 g of CrCl3? Mass of Chromium: g Cr = ? 200.0 g CrCl3 1 mol CrCl3 2 mol Cr 52.00 g Cr 200.0 g CrCl3 X ------------- X ------------- X ------------- 158.4 g CrCl3 2 mol CrCl3 1 mol Cr = 65.66 g Cr Part 1

  35. Stoichiometry 10.55 When chromium metal reacts with chlorine gas, a violet solid with the formula CrCl3 is formed: 2Cr + 3Cl2 2CrCl3. How many grams of Cr and how many grams of Cl2 are needed to produce 200.0 g of CrCl3? Mass of Chlorine: g Cl2 = ? 200.0 g CrCl3 1 mol CrCl3 3 mol Cl2 70.90 g Cl2 200.0 g CrCl3 X ------------- X ------------- X ------------- 158.4 g CrCl3 2 mol CrCl3 1 mol Cl2 = 134.3 g Cl2

  36. Stoichiometry 10.56 Black silver sulfide can be produced from the reaction of silver metal with sulfur: 2Ag + S  Ag2S. How many grams of Ag and how many grams of S are needed to produce 150.0 g of Ag2S? g Ag = ? 150.0 g Ag2S 1 mol Ag2S 2 mol Ag 107.9 g Ag 150.0 g Ag2S X ------------- X ------------- X ------------- 247.9 g Ag2S 1 mol Ag2S 1 mol Ag =130.6 g Ag Part 1

  37. Stoichiometry 10.56 Black silver sulfide can be produced from the reaction of silver metal with sulfur: 2Ag + S  Ag2S. How many grams of Ag and how many grams of S are needed to produce 150.0 g of Ag2S? g S = ? 150.0 g Ag2S 1 mol Ag2S 1 mol S 32.07 g S 150.0 g Ag2S X ------------- X ------------- X ------------- 247.9 g Ag2S 1 mol Ag2S 1 mol S =19.41 g S Part 2

  38. Limiting Reactant 10.57 What will be the limiting reactant in the production of “three nut-four bolt” combinations from a collection of 216 nuts and 284 bolts? Combinations = ? 216 nuts 284 bolts 1 combination 216 nuts X ------------- = 72 combinations 3 nuts 1 combination 284 bolts X ------------- = 71 combinations 4 bolts The limiting reactant is 284 bolts

  39. Limiting Reactant 10.58 What will be the limiting reactant in the production of “five nut-four bolt” combinations from a collection of 785 nuts and 660 bolts? Combinations = ? 785 nuts 660 bolts Nuts is the limiting reactant 1 combination 785 nuts = 157 combinations X ----------------- 5 nuts 1 combination = 165 combinations 660 bolts X ----------------- 4 bolts

  40. 10.59 A model airplane kit is designed to contain two wings, one fuselage, four engines, and six wheels. How many model airplane kits can a manufacturer produce from a parts inventory of 426 wings, 224 fuselages, 860 engines, and 1578 wheels? 426 wings, 224 fuselages, 860 engines, and 1578 wheels kits = ? Limiting Reactant 1 kit = 213 kits 426 wings X ----------------- 2 wings 1 kit 224 fuselages X ----------------- = 224 kits 1 fuselage 1 kit 860 engines =215 kits X ----------------- 4 engines 1 kit 1578 wheels X ----------------- = 263 kits 6 wheels

  41. 10.60 A model car kit is designed to contain one body, four wheels, two bumpers, and one steering wheel. How may model car kits can a manufacturer produce from a parts inventory of 137 bodies, 532 wheels, 246 bumpers, and 139 steering wheels? 137 bodies, 532 wheels, 246 bumpers, and 139 steering wheels kits = ? Limiting Reactant 1 kit = 137 kits 137 bodies X ----------------- 1 body 1 kit 532 wheels X ----------------- = 133 kits 4 wheels 1 kit 246 bumpers =123 kits X ----------------- 2 bumpers 1 kit 139 steering wheels = 139 kits X ----------------- 1 steering wheel

  42. 10.61 At high temperatures and pressures nitrogen will react with hydrogen to produce ammonia as shown by the equation N2 + 3H2  2NH3. For each of the following combinations of reactants, decide which is the limiting reactant. a) 1.25 mol N2 and 3.65 mol H2 b) 2.60 mol N2 and 8.00 mol H2 c) 44.0 N2 and 3.00 mol H2 d) 55.0 g N2 and 15.0 g H2

  43. N2 + 3H2  2NH3 Limiting Reactant 10.61 a) 1.25 mol of N2 and 3.65 mol H2 Limiting reactant = ? 1.25 mol N2 3.65 mol H2 2 mol NH3 1.25 mole N2 = 2.50 mol NH3 X ----------------- 1 mol N2 2 mol NH3 3.65 mol H2 =2.43 mol NH3 X ----------------- 3 mol H2 3.65 mol H2 is the limiting reactant.

  44. N2 + 3H2  2NH3 Limiting Reactant 10.61 b) 2.60 mol of N2 and 8.00 mol H2 Limiting reactant = ? 2.60 mol of N2 and 8.00 mol H2 2 mol NH3 2.60 mole N2 = 5.20 mol NH3 X ----------------- 1 mol N2 2 mol NH3 8.00 mol H2 =5.33 mol NH3 X ----------------- 3 mol H2 2.60 mol N2 is the limiting reactant.

  45. N2 + 3H2  2NH3 10.61 c) 44.0 g of N2 and 3.00 mol H2 Limiting reactant = ? 44.0 g of N2 and 3.00 mol H2 2 mol NH3 1 mol N2 44.0 g N2 = 3.14 mol NH3 X ----------------- X ----------------- 28.0 g N2 1 mol N2 2 mol NH3 3.00 mol H2 = 2.00 mol NH3 X ----------------- 3 mol H2 3.00 mol H2 is the limiting reactant.

  46. N2 + 3H2  2NH3 10.61 d) 55.0 g of N2 and 15.0 g H2 Limiting reactant = ? 44.0 g of N2 and 3.00 mol H2 2 mol NH3 1 mol NH3 55.0 g N2 = 3.93 mol NH3 X ----------------- X ----------------- 28.0 g N2 1 mol N2 1 mol NH3 2 mol NH3 3.00 mol H2 = 4.95 mol NH3 X ----------------- X ----------------- 3 mol H2 2.02 g H2 55.0 g N2 is the limiting reactant.

  47. Limiting Reactant 4Al + 3O2 2Al2O3

  48. 4Al + 3O2 2Al2O3 Limiting Reactant 10.6 a) 3.00 mol Al and 4.00 mol O2 Limiting reactant = ? 3.00 mol Al and 4 mol O2 2 mol Al2O3 3.00 mol Al X ----------------- =1.50 mol Al2O3 4 mol Al 2 mol Al2O3 4.00 mol O2 = 2.67 mol Al2O3 X ----------------- 3mol O2 3.00 mol Al is the limiting reactant.

  49. 4Al + 3O2 2Al2O3 10.6 b) 7.00 mol Al and 5.40 mol O2 Limiting reactant = ? 7.00 mol Al and 7.40 mol O2 2 mol Al2O3 7.00 mol Al X ----------------- =3.50 mol Al2O3 4 mol Al 2 mol Al2O3 5.40 mol O2 = 3.60 mol Al2O3 X ----------------- 3mol O2 7.00 mol Al is the limiting reactant.

  50. 4Al + 3O2 2Al2O3 10.6 c) 16.2 g Al and 0.40 mol O2 Limiting reactant = ? 16.2 g Al and 0.40 mol O2 1 mol Al 2 mol Al2O3 16.2 Al X ----------------- X ----------------- = 0.300 mol Al2O3 27.0 g Al 4 mol Al 2 mol Al2O3 0.40 mol O2 = 0.27 mol Al2O3 X ----------------- 3mol O2 0.40 mol O2 is the limiting reactant.

More Related