1. Stoker Exercises How is stoichiometry like baking cookies? 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. 10.45How 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) SiO 2 + 3C  2CO + SiC X ------------------ 60.09 g SiO 2 1 mol SiO 2 3 mol C 1 mol C 12.01 g C g C = ?1.772 g SiO 2 = 1.062 g C Stoichiometry

3. 10.45How 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 O 2 + C 3 H 8  3CO 2 + 4H 2 O X ------------------ 32.00 g O 2 1 mol O 2 5 mol O 2 1 mol C 3 H 8 44.09 g C 3 H 8 g C 3 H 8 = ?1.772 g O 2 = 0.4883 g C 3 H 8 Stoichiometry

4. 10.45How 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) CH 4 + 4Cl 2  4HCl + CCl 4 X ------------------ 16.04 g CH 4 1 mol CH 4 4 mol Cl 2 1 mol Cl 2 70.90 g Cl 2 g Cl 2 = ?1.772 g CH 4 = 31.33 g Cl 2 Stoichiometry

5. 10.45How 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) 3NO 2 + H 2 O  2HNO 3 + NO X ------------------ 46.01 g NO 2 1 mol NO 2 3 mol NO 2 1 mol H 2 O 18.02 g H 2 O g H 2 O= ?1.772 g NO 2 = 0.2313 g H 2 O Stoichiometry

6. 10.46 aHow 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) H 2 O 2 + H 2 S  2H 2 O + S X ------------------ 34.02 g H 2 O 2 1 mol H 2 O 2 1 mol H 2 S 34.09 g H 2 S g H 2 S = ?12.56 g H 2 O 2 = 12.59 g H 2 S Stoichiometry

7. 10.46 bHow 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) 4NH 3 + 3O 2  2N 2 + 6H 2 O X ------------------ 17.04 g NH 3 1 mol NH 3 4 mol NH 3 3 mol O 2 1 mol O 2 32.00 g O 2 g O 2 = ?12.56 g NH 3 = 17.69 g O 2 Stoichiometry

8. 10.46 cHow 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  MgCl 2 + H 2 X ------------------ 24.30 g Mg 1 mol Mg 2 mol HCl 1 mol HCl 36.46 g HCl g HCl = ?12.56 g Mg = 37.69 g HCl Stoichiometry

9. 10.46 dHow 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  3H 2 + 2AlCl 3 X ------------------ 36.46 g HCl 1 mol HCl 6 mol HCl 2 mol Al 1 mol Al 26.98 g Al g Al = ?12.56 g HCl = 3.098 g Al Stoichiometry

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

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

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

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

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

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

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

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

18. 2LiOH + CO 2  Li 2 CO 3 + H 2 O X ------------------ 1 mol CO 2 2 mol LiOH 1 mol LiOH 23.9 g LiOH g LiOH = ?4.50 mol CO 2 215 g LiOH 10.49 One way to remove gaseous carbon dioxide (CO 2 ) from the air in a spacecraft is to let canisters of solid lithium hydroxide (LiOH) absorb it according to the reaction 2LiOH + CO 2  Li 2 CO 3 + H 2 O. Based on this equation, calculate how many grams of LiOH must be used to achieve the following: a) absorb 4.50 moles of CO 2 Stoichiometry

19. 2LiOH + CO 2  Li 2 CO 3 + H 2 O X ------------------ 6.02 X 10 23 molec. CO 2 1 mol CO 2 2 mole LiOH g LiOH = ?3.00 X 10 24 molc. CO 2 238 g LiOH 10.49 One way to remove gaseous carbon dioxide (CO 2 ) from the air in a spacecraft is to let canisters of solid lithium hydroxide (LiOH) absorb it according to the reaction 2LiOH + CO 2  Li 2 CO 3 + H 2 O. Based on this equation, calculate how many grams of LiOH must be used to achieve the following: b) absorb 3.00 X 10 24 molec. CO 2 1 mol LiOH 23.9 g LiOH X ------------------ Stoichiometry

20. 2LiOH + CO 2  Li 2 CO 3 + H 2 O X ------------------ 18.0 g H 2 O 1 mol H 2 O 2 mole LiOH g LiOH = ?10.0 g H 2 O 26.6 g LiOH 10.49 One way to remove gaseous carbon dioxide (CO 2 ) from the air in a spacecraft is to let canisters of solid lithium hydroxide (LiOH) absorb it according to the reaction 2LiOH + CO 2  Li 2 CO 3 + H 2 O. Based on this equation, calculate how many grams of LiOH must be used to achieve the following: c) produce 10.0 g H 2 O X ------------------ 1 mol LiOH 23.9 g LiOH Stoichiometry

21. 2LiOH + CO 2  Li 2 CO 3 + H 2 O X ------------------ 73.9 g Li 2 CO 3 1 mol Li 2 CO 3 2 mole LiOH g LiOH = ?10.0 g Li 2 CO 3 6.47 g LiOH 10.49 One way to remove gaseous carbon dioxide (CO 2 ) from the air in a spacecraft is to let canisters of solid lithium hydroxide (LiOH) absorb it according to the reaction 2LiOH + CO 2  Li 2 CO 3 + H 2 O. Based on this equation, calculate how many grams of LiOH must be used to achieve the following: d) produce 10.0 g of Li 2 CO 3 X ------------------ 1 mol LiOH 23.9 g LiOH Stoichiometry

22. 10.50 a) Tungsten (W) metal, used to make incandescent light bulb filaments, is produced by the reaction WO 3 (s) + 3H 2 (g)  W(s) + 3H 2 O(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 10 9 molec. H 2 O b) 2.53 moles of H 2 Od) 250 000 atoms of W X ------------------ 183.8 g W 1 mol W 1 mol WO 3 g WO 3 = ?10.00 g W X ------------------ 1 mol WO 3 231.8 g WO 3 WO 3 (s) + 3H 2 (g)  W(s) + 3H 2 O(l) 12.61 g WO 3 Stoichiometry

23. 10.50 b) Tungsten (W) metal, used to make incandescent light bulb filaments, is produced by the reaction WO 3 (s) + 3H 2 (g)  W(s) + 3H 2 O(l). Based on this equation, calculate how many grams of WO 3 are needed to produce each of the following: a) 10.00 g of W b) 1.00 X 10 9 molec. H 2 O c) 2.53 moles of H 2 Od) 250 000 atoms of W X ------------------ 6.02 X 10 23 molec. H 2 O 1 mol H 2 O 3 mol H 2 O 1 mol WO 3 g WO 3 = ?1.00 X 10 9 molec. H 2 O X ------------------ 1 mol WO 3 231.8 g WO 3 WO 3 (s) + 3H 2 (g)  W(s) + 3H 2 O(l) 1.28 X 10 -13 g WO 3 Stoichiometry

24. 10.50 c) Tungsten (W) metal, used to make incandescent light bulb filaments, is produced by the reaction WO 3 (s) + 3H 2 (g)  W(s) + 3H 2 O(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 10 9 molec. H 2 O c) 2.53 moles of H 2 Od) 250 000 atoms of W X ------------------ 3 mol H 2 O 1 mol WO 3 231.8 g WO 3 g WO 3 = ?2.53 mol H 2 O WO 3 (s) + 3H 2 (g)  W(s) + 3H 2 O(l) = 195 g WO 3 Stoichiometry

25. 10.50 d) Tungsten (W) metal, used to make incandescent light bulb filaments, is produced by the reaction WO 3 (s) + 3H 2 (g)  W(s) + 3H 2 O(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 10 9 molec. H 2 O c) 2.53 moles of H 2 Od) 250 000 atoms of W X ------------------ 6.0 X 10 23 atoms W 1 mol W 1 mol WO 3 g WO 3 = ?2.5 X 10 5 atoms W WO 3 (s) + 3H 2 (g)  W(s) + 3H 2 O(l) = 9.7 X 10 -17 g WO 3 X ------------------ 1 mol WO 3 231.8 g WO 3 Stoichiometry

28. X ------------------ 36.46 g HCl 1 mol HCl 8 mol HCl 1 mol K 2 S 2 O 3 mol K 2 S 2 O 3 = ?2.500 g HCl = 0.008571 mol K 2 S 2 O K 2 S 2 O 3 + 4Cl 2 + 5H 2 O  2 KHSO 4 + 8HCl Stoichiometry

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

33. 10.54 g Be = ?45.0 g N 2 How many grams of beryllium are needed to react completely with 45.0 g of nitrogen in the synthesis of Be 3 N 2 ? 3Be + N 2  Be 3 N 2 45.0 g N 2 X ------------- 28.0 g N 2 1 mol N 2 X ------------- 1 mol N 2 3 mol Be X ------------- 1 mol Be 9.01 g Be = 43.4 g Be Stoichiometry

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

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

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

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

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

39. Nuts is the 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 785 nuts X ----------------- 1 combination 5 nuts = 157 combinations 660 boltsX ----------------- 1 combination 4 bolts = 165 combinations Limiting Reactant

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? kits = ? 426 wings, 224 fuselages, 860 engines, and 1578 wheels 426 wingsX ----------------- 1 kit 2 wings = 213 kits 224 fuselagesX ----------------- 1 kit 1 fuselage = 224 kits 860 engines X ----------------- 1 kit 4 engines =215 kits 1578 wheelsX ----------------- 1 kit 6 wheels = 263 kits Limiting Reactant

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? kits = ? 137 bodies, 532 wheels, 246 bumpers, and 139 steering wheels 137 bodiesX ----------------- 1 kit 1 body = 137 kits 532 wheelsX ----------------- 1 kit 4 wheels = 133 kits 246 bumpers X ----------------- 1 kit 2 bumpers =123 kits 139 steering wheels X ----------------- 1 kit 1 steering wheel = 139 kits Limiting Reactant

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

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

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

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

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

47. 4Al + 3O 2  2Al 2 O 3 Limiting Reactant

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

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

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

51. Limiting reactant = ? 100.0 g Al and 100.0 g O 2 100.0 g Al X ----------------- 2 mol Al 2 O 3 4 mol Al = 1.852 mol Al 2 O 3 100.0 g O 2 X ----------------- 1 mol Al 2 O 3 32.00 g O 2 = 2.083 mol Al 2 O 3 100.0 g Al is the limiting reactant. 10.6 d) 100.0 g Al and 100.0 g O 2 4Al + 3O 2  2Al 2 O 3 27.00 g Al 1 mol Al X ----------------- 3 mol O 2 2 mol Al 2 O 3

52. How many grams of magnesium nitride can be produced from the following amounts of reactants? a) 10.0 g of Mg and 10.0 g of N 2 Limiting Reactant

53. 10.63 a) Determine the limiting reactant in each part, then convert moles of product to mass of product. How many grams of magnesium nitride can be produced from the following amounts of reactants? a) 10.0 g of Mg and 10.0 g of N 2 g of Mg 3 N 2 = ? 10.0 g of Mg and 10.0 g of N 2 10.0g Mg X ---------------- 24.3 g Mg 1 mol Mg X ---------------- 3 mol Mg 1 mol Mg 3 N 2 = 0.137 ml Mg 3 N 2 28.0 g N2 1 mol N 2 X ---------------- 1 mol N 2 1 mol Mg 3 N 2 10.0 g N 2 X -------------- = 0.357 mol Mg 3 N 2 The Mg is the limiting reactant. 0.137 mol Mg 3 N 2 X ------------------- 1 mol Mg 3 N 2 100.9 g Mg 3 N 2 = 13.8 g Mg 3 N 2 Limiting Reactant

54. 10.63 b) 20.0 g Mg X --------------------------- 1 mol Mg 24.3 g Mg X ------------------------ 3 mol Mg 1 mol Mg 3 N 2 = 0.274 mol Mg 3 N 2 28.0 g N2 1 mol N 2 X ---------------- 1 mol N 2 1 mol Mg 3 N 2 10.0 g N 2 X -------------- = 0.357 mol Mg 3 N 2 0.274 mol Mg 3 N 2 X ----------------------- 100.9 g Mg 3 N 2 1 mol Mg 3 N 2 = 27.6 g Mg 3 N 2 Mg is the limiting reactant. Limiting Reactant

55. 10.63 c) 30.0 g Mg X --------------------------- 1 mol Mg 24.3 g Mg X ------------------------ 3 mol Mg 1 mol Mg 3 N 2 = 0.412 mol Mg 3 N 2 28.0 g N 2 1 mol N 2 X ---------------- 1 mol N 2 1 mol Mg 3 N 2 10.0 g N 2 X -------------- = 0.357 mol Mg 3 N 2 0.357 mol Mg 3 N 2 X ----------------------- 100.9 g Mg 3 N 2 1 mol Mg 3 N 2 = 36.0 g Mg 3 N 2 N 2 is the limiting reactant. Limiting Reactant

56. 10.63 d) 40.0 g Mg X --------------------------- 1 mol Mg 24.3 g Mg X ------------------------ 3 mol Mg 1 mol Mg 3 N 2 = 0.549 mol Mg 3 N 2 28.0 g N 2 1 mol N 2 X ---------------- 1 mol N 2 1 mol Mg 3 N 2 10.0 g N 2 X -------------- = 0.357 mol Mg 3 N 2 0.357 mol Mg 3 N 2 X ----------------------- 100.9 g Mg 3 N 2 1 mol Mg 3 N 2 = 36.0 g Mg 3 N 2 N 2 is the limiting reactant. Limiting Reactant

57. 70.0 g Fe 3 O 4 12.0 g of oxygen gas sealed in a glass vessel. Unreacted reactant = ?70.0 g Fe 3 O 4 12 g O 2 The limiting reactant will all react. Use the limiting reactant to determine the mass of the other reactant that reacts, and then find the unreacted by difference. Limiting Reactant

58. 0.455 mol Fe 2 O 3 2.25 mol Fe 2 O 3 Iron(III) oxide 70.0 g Fe3O4X -------------------------- - 1 mol Fe3O4 231. g Fe3O4 X ------------------------ 4 mol Fe3O4 6 mole Fe2O3 = 0.455 mole Fe 2 O 3 32.0 g O 2 1 mol O 2 X ---------------- 1 mol O 2 6 mol Fe 2 O 3 12.0 g O 2 X -------------- = 2.25 mol Fe 2 O 3 The Fe 3 O 4 is the limiting reactant. There will be none left upon completion. Calculate the mass of O 2 reacted: Limiting Reactant

59. 0.455 mol Fe 2 O 3 X --------------------------- 1 mol O 2 6 mol Fe 2 O 3 X ------------------------ 1 mol O 2 32.0 g O 2 = 2.43 g O 2 Limiting Reactant

61. 16.0 10.1 Limiting Reactant

62. g of products = ? 8.00 g SCl 2 and 4.00 g NaF

63. g of Products = ? 100.00 g Fe 3 Br 8 300.0 g Fe 3 Br 8 Na 2 CO 3 is the limiting reactant. Limiting Reactant 10.70 Determine the number of grams of each of the products that can be made from 100.00 g Na 2 CO 3 and 300.0 g Fe 3 Br 8 by the following reaction. 4Na 2 CO 3 + Fe 3 Br 8  8NaBr + 4CO 2 + Fe3O 4

64. Calculation of Theoretical Yield The theoretical yield of a reaction is the amount of product that would be formed if the reaction went to completion. It is based on the stoichiometry of the reaction and ideal conditions in which starting material is completely consumed, undesired side reactions do not occur, the reverse reaction does not occur, and there no losses in the work-up procedure. Theoretical YieldActual Yield and Percent Yield

65. After your laboratory reaction is complete, you will isolate and measure the amount of product, then compare the actual yield to the theoretical yield to determine the percent yield: Actual yield (in grams) -------------------------------- = X 100% = percent yield Theoretical yield (in grams) Theoretical YieldActual Yield and Percent Yield

66. (a) What is the theoretical yield of NaClO that can be obtained from a reaction mixture containing 75.0 g of NaOH and 50.0g of Cl 2 ? (b) If the actual yield of NaClO for the reaction mixture in part (a) is 43.2 g, what is the percent yield of NaClO for the reaction? Example 10.13 The active ingredient in household laundry bleaches is sodium hypochlorite (NaClO). 2NaOH + Cl 2  NaCl + NaClO + H 2 O Theoretical YieldActual Yield and Percent Yield

67. Answer for Example 10.13 (a)The active ingredient in household laundry bleaches is sodium hypochlorite (NaClO). 2NaOH + Cl 2  NaCl + NaClO + H 2 O 75.0 g NaOH X --------------------------- 1 mol NaOH 40.0 g NaOH X ------------------------ 2 mol NaOH 1 mol NaClO = 0.938 mol NaClO 71.0 g Cl 2 1 mol Cl 2 X ---------------- 1 mol Cl 2 1 mol NaClO =0.704 mol NaClO The calculations show that Cl 2 is the limiting reactant. The maximum number of grams of NaClO obtainable from the limiting reactant, that is, the theoretical yield, can now be calculated. It is done using a one-step “moles of A” to “grams of “A” setup. 50.0 g Cl 2 X --------------------------- Theoretical YieldActual Yield and Percent Yield

68. 0.704 mol NaClO X ----------------------- 75.5 g NaClO 1 mol NaClO = 52.4 g NaClO Part b  Theoretical YieldActual Yield and Percent Yield

69. (b) If the actual yield of NaClO for the reaction mixture in part (a) is 43.2 g, what is the percent yield of NaClO for the reaction? Example 10.13 The active ingredient in household laundry bleaches is sodium hypochlorite (NaClO). 2NaOH + Cl 2  NaCl + NaClO + H 2 O actual yield Percent yield = ----------------------- X 100 = theoretical yield % yield = -------------------------- 52.4 g NaClO 43.2 g NaClO X 100 =82.4 % NaClO Actual Yield and Percent Yield

70. % yield = 16.0 g (actual) 52.0 g (theoretical) X 100 =30.8 % % yield = ? 16.0 g product 52.0 g th. yield Theoretical and Percent Yield 10.71

71. % yield = 24.79 g (actual) 25.31 g (theoretical) X 100 =97.95 % % yield = ? Th. Yield 25.31 and product 24.79 g Theoretical and Percent Yield 10.72

72. 75.0 g Al X --------------------------- 1 mol Al 27.0 g Al X ------------------------ 2 mol Al 1 mol Al 2 S 3 = 1.39 mol Al 2 S 3 32.07 g S 1 mol S X ---------------- 3 mol S 1 mol Al 2 S 3 = 3.118 mol Al 2 S 3 300.0 g S X --------------------------- Al is the limiting reactant. Theoretical and Percent Yield 10.73

73. 1.39 mol Al 2 S 3 X -------------------------- 150. g Al 2 S 3 1 mol Al 2 S 3 = 209 g Al 2 S 3 % yield = ------------------------------ 125 g Al 2 S 3 (actual) 209 g Al 2 S 3 (theor) 59.8 % Theoretical Yield X 100 = Theoretical and Percent Yield

74. Th. Yield = ? 75.0 g Aland 200.0 g O 2 Theoretical and Percent Yield

75. 1.39 mol Al 2 O 3 X -------------------------- 102 g Al 2 O 3 1 mol Al 2 O 3 = 142 g Al 2 O 3 % yield = ------------------------------ 125 g Al 2 O 3 (actual) 142 g Al 2 O 3 (theor) 88.0 % Theoretical Yield X 100 = Theoretical and Actual yield

76. % yield = ? 2.130 g H 2 2.130 g H 2 X --------------------------- 1 mol H 2 2.016 g H 2 X -------------------- 1 mol H 2 2 mol HCl = 96.44 % HCl ---------------- 77.04 g HCl 74.30 g HCl X 100 % yield = X --------------- 1 mol HCl 36.46 g HCl = 77.04 g HCl Step 2: Find the percent yield of HCl Step 1: We know from the question that H 2 is the limiting reaction. We need to find how many grams of HCl can be produced from 2.130 g H 2. 10.75 If 74.30 g of HCl were produced from 2.130 g of H 2 and an excess of Cl 2 according to reaction H 2 + Cl 2  2HCl, what was the percent yield of HCl? ---------------- Theoretical yield Actual yield % yield = X 100 Theoretical and Percent Yield

77. % yield = ? 28.2 g N 2 28.2 g N 2 X -------------------------- 1 mol N 2 28.0 g N 2 X ------------------- 1 mol Ca 3 N 2 1 mol N 2 = 149 g Ca 3 N 2 % Yield = X ----------------- 1 mol Ca 3 N 2 148 g Ca 3 N 2 X 100 = 149 g Ca 3 N 2 115.7 g Ca 3 N 2 ---------------------77.7% Theoretical and Actual yield

78. Actual yield = ? 55.0 g Al 55.0 g Al X -------------------------- 1 mol Al 27.0 g Al X ------------------- 1 mol Al 2 S 3 2 mol Al = 153 g Al 2 S 3 153 g Al 2 S 3 (theor) X ----------------- 1 mol Al 2 S 3 150. g Al 2 S 3 X 100 = 100 Al 2 S 3 (theor) 85.6 g Al 2 S 3 (actual) ---------------------131 g Al 2 S 3 Theoretical yield of Al 2 S 3 Actual yield = theoretical yield X % yield Theoretical and Actual yield 10.77

79. Actual yield = ? 75.0 g Ag 75.0 g Ag X -------------------------- 1 mol Ag 107.9 g Ag X ------------------- 1 mol Ag 2 S 2 mol Ag = 86.2 g Ag 2 S 86.2 g Ag 2 S (theor) X ----------------- 1 mol Ag 2 S 248 g Ag 2 S X 100 = 100 Ag 2 S (theor) 72.9 g Ag 2 S (actual) ---------------------62.8 g Ag 2 S Theoretical yield of Ag 2 S Actual yield = theoretical yield X % yield Theoretical and Actual yield 10.78

80. g of product = ? 35.0 g CO and O 2 35.0 g CO X --------------------------- 1 mol CO 28.0 g CO X ------------------------ 2 mol CO 2 mol CO 2 = 1.25 mol CO 2 32.0 g O 2 1 mol O2 X ---------------- 1 mol O 2 2 mol CO 2 =2.19 mol CO 2 35.0 g O 2 X --------------------------- The CO is the limiting reactant. Theoretical and Actual yield 10.79

81. 1.25 mol CO 2 X --------------------------- 44.0 g CO 2 1 mol CO 2 = 55.0 g CO 2 55.0 g CO 2 (theor) X -------------------- 44.0 g CO 2 (actual) 100 g CO 2 (theor) = 31.8 g CO 2 Theoretical Yield Actual yield Theoretical and Actual yield

82. g of product = ? 2.25 of C and O 2 Theoretical and Actual yield 10.80

83. Simultaneous Reactions

84. A X --------------------------- C B X ------------------------ D E = F B C X ---------------- D E = FA X ---------------------------