1. Limiting Reagent Problems Video Example

2. Limiting Reagent Problems Video Example Reagent = Reactant

3. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation

5. a)Determine the excess and limiting reagents.

6. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. b)What mass of AlBr 3 can be produced?

7. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. b)What mass of AlBr 3 can be produced? c)What mass of the excess reagent is left over?

8. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

9. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

10. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

11. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

12. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

13. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

14. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

15. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

16. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

17. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially. 1.85 mol Al present 2.50 mol Br 2 present

18. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts. 1.85 mol Al present 2.50 mol Br 2 present

19. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts. 1.85 mol Al present 2.50 mol Br 2 present

20. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts. 1.85 mol Al present 2.50 mol Br 2 present

21. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts. 1.85 mol Al present 2.50 mol Br 2 present

22. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts. 1.85 mol Al present 2.50 mol Br 2 present

23. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts. 1.85 mol Al present 2.50 mol Br 2 present

24. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts. 1.85 mol Al present 2.50 mol Br 2 present 2.78 mol Br 2 required

25. 2.78 mol Br 2 required 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts. 1.85 mol Al present 2.50 mol Br 2 present 2.78 mol Br 2 required

26. 2.78 mol Br 2 required 2.50 mol Br 2 present 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts. 1.85 mol Al present 2.50 mol Br 2 present 2.78 mol Br 2 required

27. 2.78 mol Br 2 required 2.50 mol Br 2 present So Br 2 is the limiting reagent. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts. 1.85 mol Al present 2.50 mol Br 2 present 2.78 mol Br 2 required

28. 2.78 mol Br 2 required 2.50 mol Br 2 present So Br 2 is the limiting reagent. And the Al is the excess reagent. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts. 1.85 mol Al present 2.50 mol Br 2 present 2.78 mol Br 2 required

29. 2.78 mol Br 2 required 2.50 mol Br 2 present So Br 2 is the limiting reagent. And the Al is the excess reagent. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. 1.85 mol Al present 2.50 mol Br 2 present The moles of substances used up or produced must be based on the moles of the limiting reagent, Br 2, present

30. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. b)What mass of AlBr 3 can be produced? c)What mass of the excess reagent is left over?

31. b)What mass of AlBr 3 can be produced? c)What mass of the excess reagent is left over? Initial moles Change in moles Final moles An ICF chart

32. b)What mass of AlBr 3 can be produced? c)What mass of the excess reagent is left over? Initial moles Change in moles Final moles

33. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation How many moles of AlBr 3 can be produced? 1.85 mol Al present 2.50 mol Br 2 present

34. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced. 1.85 mol Al present 2.50 mol Br 2 present How many moles of AlBr 3 can be produced?

35. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced. 1.85 mol Al present 2.50 mol Br 2 present The moles of substances used up or produced must be based on the moles of the limiting reagent, Br 2, present How many moles of AlBr 3 can be produced?

36. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced. 1.85 mol Al present 2.50 mol Br 2 present The moles of substances used up or produced must be based on the moles of the limiting reagent, Br 2, present How many moles of AlBr 3 can be produced?

37. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced. 1.85 mol Al present 2.50 mol Br 2 present Limiting Reagent How many moles of AlBr 3 can be produced?

38. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced. 1.85 mol Al present 2.50 mol Br 2 present How many moles of AlBr 3 can be produced?

39. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced. 1.85 mol Al present 2.50 mol Br 2 present How many moles of AlBr 3 can be produced?

40. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced. 1.85 mol Al present 2.50 mole Br 2 present How many moles of AlBr 3 can be produced?

41. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced How many moles of AlBr 3 can be produced?

42. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation How many moles of Al are used up? Step 4– Determine the moles of Al used up. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced

43. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 4– Determine the moles of Al used up. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Limiting Reagent How many moles of Al are used up?

44. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 4– Determine the moles of Al used up. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Limiting Reagent How many moles of Al are used up?

45. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 4– Determine the moles of Al used up. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced How many moles of Al are used up?

46. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 4– Determine the moles of Al used up. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced How many moles of Al are used up?

47. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 4– Determine the moles of Al used up. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced

48. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 4– Determine the moles of Al used up. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced How many moles of Al are used up?

49. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles Final moles

50. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles Final moles Excess Reagent

51. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles Final moles Excess Reagent Limiting Reagent

52. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles1.85 Change in moles Final moles Excess Reagent Limiting Reagent

53. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles1.852.50 Change in moles Final moles Excess Reagent Limiting Reagent

54. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles1.852.500 Change in moles Final moles Excess Reagent Limiting Reagent

55. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles1.852.500 Change in moles –1.67 Final moles Excess Reagent Limiting Reagent

56. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles1.852.500 Change in moles –1.67–2.50 Final moles Excess Reagent Limiting Reagent

57. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles1.852.500 Change in moles –1.67–2.50+1.67 Final moles

58. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles1.852.500 Change in moles –1.67–2.50+1.67 Final moles0.18

59. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles1.852.500 Change in moles –1.67–2.50+1.67 Final moles0.180

60. 1.67 mol Al used up 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances. 1.85 mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles1.852.500 Change in moles –1.67–2.50+1.67 Final moles0.1801.67

61. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced.

62. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced.

63. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced.

64. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced.

65. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced.

66. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced.

67. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced. 445 g of AlBr 3 is produced

68. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? 445 g of AlBr 3 is produced

69. 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

70. 445 g of AlBr 3 is produced 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

71. 445 g of AlBr 3 is produced 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

72. 445 g of AlBr 3 is produced 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

73. 445 g of AlBr 3 is produced 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

74. 445 g of AlBr 3 is produced 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

75. 445 g of AlBr 3 is produced 4.86 g of the excess reagent, Al, is left over 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

76. 445 g of AlBr 3 is produced 4.86 g of the excess reagent, Al, is left over When 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation

77. 445 g of AlBr 3 is produced 4.86 g of the excess reagent, Al, is left over When 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Al is the excess reagent

78. 445 g of AlBr 3 is produced 4.86 g of the excess reagent, Al, is left over When 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Al is the excess reagent and Br 2 is the limiting reagent.

79. 445 g of AlBr 3 is produced 4.86 g of the excess reagent, Al, is left over When 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Al is the excess reagent and Br 2 is the limiting reagent. b)445 g of AlBr 3 is produced.

80. 445 g of AlBr 3 is produced 4.86 g of the excess reagent, Al, is left over When 50.0 g of Al (s) is mixed with 400.0 g of Br 2(l) and allowed to react according to the balanced equation a)Al is the excess reagent and Br 2 is the limiting reagent. b)445 g of AlBr 3 is produced. c)4.86 g of Al, the excess reagent, is left over after the reaction.