3. Limiting Reagent u If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make?

4. Limiting Reagent u The limiting reagent is the reactant you run out of first. u The excess reagent is the one you have left over. u The limiting reagent determines how much product you can make.

5. Determining the Limiting Reactant u Do two stoichiometry problems. u The one that makes the least product is the limiting reagent.

6. Example u Copper reacts with sulfur to form copper (I) sulfide. If 10.6 g of copper reacts with 3.83 g sulfur, how much product will be formed?

7.  2 Cu + S  Cu 2 S 10.6 g Cu 63.5 g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S 159.1 g Cu 2 S = 13.3 g Cu 2 S Example

8.  2 Cu + S  Cu 2 S 3.83 g S 32.1 g S 1 mol S 1 S 1 Cu 2 S 1 mol Cu 2 S 159.1 g Cu 2 S = 19.0 g Cu 2 S Example

9.  2 Cu + S  Cu 2 S Example u 10.6 grams of copper resulted in 13.3 grams of copper (I) sulfide. u 3.83 grams of sulfur resulted in 19.0 grams of copper (I) sulfide.

10.  2 Cu + S  Cu 2 S Cu is the limiting reactant! Example u Using 10.6 grams of copper resulted in the smaller amount of copper (I) sulfide formed. Therefore

11.  2 Cu + S  Cu 2 S Example u Considering the limiting reactant, 13.3 grams of copper (I) sulfide are formed.

12. Example u Determine the mass of P 4 O 10 formed if 25.0 g of P 4 and 50.0 g of O 2 are combined. Which reactant is the limiting reactant?

13. u Write the balanced equation  P 4 + 5 O 2  P 4 O 10 Example

14.  P 4 + 5 O 2  P 4 O 10 25.0 g P4P4 124.0 P4P4 1 mol P4P4 1 P4P4 1 P 4 O 10 1 mol P 4 O 10 284.0 g P 4 O 10 = 57.3 g P 4 O 10 Example

15.  P 4 + 5 O 2  P 4 O 10 50.0 g O2O2 32.0 O2O2 1 mol O2O2 5 O2O2 1 P 4 O 10 1 mol P 4 O 10 284.0 g P 4 O 10 = 88.8 g P 4 O 10 Example

16. u P 4 is the limiting reactant, and 57.3 grams of P 4 O 10 are formed.

17. Problem u The reaction between solid sodium and iron (III) oxide is one in a series of reactions that inflates an automobile airbag. If 100. g of Na and 100. g of Fe 2 O 3 are used in the reaction, determine the limiting reactant and the mass of solid iron produced. Fe 2 O 3 limits, 69.9 g of Fe produced

18. Problem u How many moles of cesium xenon heptafluoride (CsXeF 7 ) can be produced from the reaction of 12.5 mol of cesium fluoride with 10.0 mol of xenon hexafluoride? 10.0 mol

19. Problem u The reaction of chlorine gas with solid phosphorus (P 4 ) produces solid PCl 5. When 22.0 g of chlorine reacts with 25.5 g of P 4, which reactant limits the amount of product? Cl 2 is limiting.

20. How Much Excess Reagent? u Convert the limiting reagent into the excess reagent. u Subtract that from the amount of excess you started with.

21. Example  Mg(s) + 2 HCl(g)  MgCl 2 (s) + H 2 (g) u If 10.1 mol of magnesium and 4.87 mol of HCl gas are reacted, how many moles excess reagent remain?

22. Example  Mg(s) + 2 HCl(g)  MgCl 2 (s) + H 2 (g) u First, determine which reactant is in excess.

23.  Mg(s) + 2 HCl(g)  MgCl 2 (s) + H 2 (g) 10.1 mol Mg 1 mol Mg 1 mol MgCl 2 = 10.1 mol MgCl 2 Example

24.  Mg(s) + 2 HCl(g)  MgCl 2 (s) + H 2 (g) 4.87 mol HCl 2 mol HCl 1 mol MgCl 2 = 2.44 mol MgCl 2 Example

25. u Mg is in excess. u Now determine how much excess! u Convert the limiting reactant to excess reactant.

26.  Mg(s) + 2 HCl(g)  MgCl 2 (s) + H 2 (g) 4.87 mol HCl 2 mol HCl 1 mol Mg = 2.44 mol Mg Example

27. u Now take the excess reactant and subtract the liming converted into excess. u 10.1 mol Mg – 2.44 mol Mg = 7.66 mol of Mg are in excess

28. Example u If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much excess reagent will remain?

29. Example u Write the balanced equation. u If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much excess reagent will remain? u 2Al + 3CuSO 4  Al 2 (SO 4 ) 3 + 3Cu

30. Example u 2Al + 3CuSO 4  Al 2 (SO 4 ) 3 + 3Cu First, determine which reactant is in excess.

31. u 2Al + 3CuSO 4  Al 2 (SO 4 ) 3 + 3Cu 10.3 g Al 27.0 g Al 1 mol Al 2 mol Al 3 mol Cu = 0.572 mol Cu Example

32. u 2Al + 3CuSO 4  Al 2 (SO 4 ) 3 + 3Cu 51.7 g CuSO 4 159.6 g CuSO4 1 mol CuSO 4 3 mol CuSO 4 3 mol Cu = 0.324 mol Cu Example

33. u Al is in excess. u Now determine how much excess! u Convert the limiting reactant to excess reactant.

34. u 2Al + 3CuSO 4  Al 2 (SO 4 ) 3 + 3Cu 51.7 g CuSO 4 159.6 CuSO 4 1 mol CuSO 4 3 mol CuSO 4 2 mol Al 1 mol Al 27.0 g Al = 5.83 g Al Example

35. u Now take the excess reactant and subtract the liming converted into excess. u 10.3 g Al – 5.83 g Al = 4.47 g of Al are in excess

36. Problem u Zn + 2MnO 2 + H 2 O  Zn(OH) 2 + Mn 2 O 3 u If 25.0 g of zinc and 39.0 g of manganese dioxide are used, how much excess reagent will remain? (Answer: 10.3 g Zn)

37. Problem u Lithium reacts spontaneously with bromine to produce lithium bromide. If 3.40 g of lithium react with 32.0 g of bromine, how much excess reactant remains? (Answer: 0.64 g Li)

38. Problem u CH 4 + 3Cl 2  CH 3 Cl + 3HCl u If 50.0 g of methane reacts with 9.67 moles of chlorine gas, how mass of excess reagent will remain? (Answer: 20.6 g of Cl 2 )

40. Yield u The yield is the amount of product made in a chemical reaction. u There are three types of yield that can be determined.

41. Yield u Actual yield - what you get in the lab when the chemicals are mixed u Theoretical yield - what the balanced equation tells you you should make Actual u Percent yield u Percent yield = x 100 % Theoretical

42. Percent Yield u Percent yield tells us how “efficient” a reaction is. u Percent yield cannot be larger than 100 %.

43. Example u 6.78 g of copper are produced when 3.92 g of Al are reacted with excess copper (II) sulfate.  2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu u What is the actual yield of copper? (Answer: 6.78 g)

44. Example u 6.78 g of copper are produced when 3.92 g of Al are reacted with excess copper (II) sulfate.  2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu u What is the theoretical yield?

45.  2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu 3.92 g Al 27.0 Al 1 mol Al 2 mol Al 3 mol Cu 1 mol Cu 63.5 g Cu = 13.8 g Cu Example

46. u 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.  2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu u What is the percent yield?

47. Yield Actual u Percent yield u Percent yield = x 100 % Theoretical 6.78 g u Percent yield u Percent yield = x 100 % 13.8 g = 49.1%

48. Problem u 58.2 g of zinc are produced when 14.7 g of Li are reacted with excess zinc sulfate.  2Li + ZnSO 4  Li 2 SO 4 + Zn u What is the actual yield of zinc? (Answer: 58.2 g)

49. Problem u 58.2 g of zinc are produced when 14.7 g of Li are reacted with excess zinc sulfate.  2Li + ZnSO 4  Li 2 SO 4 + Zn u What is the theoretical yield of zinc? (Answer: 69.7 g)

50. Problem u 58.2 g of zinc are produced when 14.7 g of Li are reacted with excess zinc sulfate.  2Li + ZnSO 4  Li 2 SO 4 + Zn u What is the percent yield of zinc? (Answer: 83.5%)