1. Basic Stoichiometry Pisgah High School M. Jones Revision history: 5/16/03, 02/04/12, 04/27/12 With the Gas Laws

2. The word stoichiometry comes from the Greek words stoicheion which means “element” and metron which means “measure”.

3. Stoichiometry deals with the amounts of reactants and products in a chemical reaction.

4. Stoichiometry deals with moles.

5. Recall that … 1 mole = 22.4 L of any gas at STP 1 mole = the molar mass molecules 1 mole = 6.022 x 10 23 atoms or

6. The word mole is one that represents a very large number. … “mole” means 6.022 x 10 23 Much like “dozen” means 12,

7. The key to doing stoichiometry is the balanced chemical equation. 2 H 2 + O 2  2 H 2 O 22

8. The coefficients give the relative number of atoms or molecules of each reactant or product … as well as the number of moles. 2 H 2 + O 2  2 H 2 O

9. 2 molecules of hydrogen 1 molecule of oxygen 2 molecules of water Two molecules of hydrogen combine with one molecule of oxygen to make two molecules of water.

10. 2 H 2 + O 2  2 H 2 O The balanced chemical equation also gives the smallest integer number of moles. 2 molecules of hydrogen 1 molecule of oxygen 2 molecules of water

11. 2 H 2 + O 2  2 H 2 O The balanced chemical equation also gives the smallest integer number of moles. 2 moles of hydrogen 1 mole of oxygen 2 moles of water

12. 2 H 2 + O 2  2 H 2 O 2 moles of hydrogen 1 mole of oxygen 2 moles of water Two moles of hydrogen combine with one mole of oxygen to make two moles of water.

13. Applications of Gay-Lussac’s Law

14. Simply put, Gay-Lussac’s Law says this: The volumes of the gases are in the same ratio as the coefficients in the balanced equation. 2 H 2 + O 2  2 H 2 O 2 moles1 mole2 moles 2 L1 L2 L

15. C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g ) 1 mol 5 mols3 mols4 mols Suppose 3.5 L of propane gas at STP is burned in oxygen, how many L at STP of oxygen will be required? 17.5 L O 2 1 L 5 L3 L4 L

16. C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g ) 1 mol 5 mols3 mols4 mols Suppose 3.5 L of propane gas at STP is burned in oxygen, how many L at STP of oxygen will be required? 17.5 L O 2 How many liters of carbon dioxide gas and water vapor at STP would be produced? 10.5 L CO 2 and 14.0 L H 2 O 1 L 5 L3 L4 L

17. CH 4 (g) + 4 Cl 2 (g)  CCl 4 (g) + 4 HCl(g) When methane gas is allowed to react with an excess of chlorine gas, tetrachloromethane and hydrogen chloride gas will be produced. How many L of methane will react with 0.800 L of chlorine gas at STP? 0.200 L Cl 2

18. Stoichiometry problems involving gases

19. Find the mass of HOCl that can be produced when 2.80 L of chlorine gas at STP reacts with excess hydrogen peroxide. Cl 2 (g) + H 2 O 2 (l)  2 HOCl (l) 2.80 L??? g Convert 2.80 L of Cl 2 gas at STP to moles

20. Cl 2 (g) + H 2 O 2 (l)  2 HOCl (l) 2.80 L??? g 0.125 mol Cl 2 0.125 mol.125 x 2 0.250 mol 13.1g HOCl

21. The reaction between copper and nitric acid

22. Will copper dissolve in acids? Cu + 2HCl  CuCl 2 + 2H 2 (g) No Reaction Most metals react with HCl to produce a metal chloride solution and H 2 gas. Not copper Consider hydrochloric acid

23. Will copper dissolve in acids? Cu + 2HCl  CuCl 2 + 2H 2 (g) No Reaction Consider hydrochloric acid Copper is below hydrogen in the activity series. Copper metal will only replace elements that are below it in the activity series.

24. What about other acids? The same is true for all acids except nitric acid Cu + HBr  NR Cu + HI  NR Cu + HF  NR Cu + H 2 SO 4  NR Cu + HC 2 H 3 O 2  NR

25. Nitric acid is the only acid that will dissolve copper. 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O A beaker contains a penny and some nitric acid is added.

26. Nitric acid is the only acid that will dissolve copper. The penny begins to disappear and the solution turns blue-green and a brown gas is given off. 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O

27. Nitric acid is the only acid that will dissolve copper. The gas produced in the reaction is NO, which is colorless. Reddish brown NO 2 forms when NO reacts with the oxygen in the air. 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O

28. Nitric acid is the only acid that will dissolve copper. The penny is gone and the solution turns a dark blue. The brown NO 2 gas escapes from the open beaker. 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O

29. Nitric acid is the only acid that will dissolve copper. Calculate the volume of NO gas at STP when 20.0 grams of copper dissolves. 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O

30. Nitric acid is the only acid that will dissolve copper. 20.0 g??? L 0.315 mol Cu 0.315 mol 0.210 mol x 0.667 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O

31. Nitric acid is the only acid that will dissolve copper. 20.0 g 4.70 L NO 0.315 mol ??? L 0.210 mol 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O x 0.667

32. Using the Combined Gas Law

33. What if the conditions are not at STP? What will be the volume of the NO gas at room temperature in the mountains? The temperature is 25 C, and the pressure is 691 torr.

34. What if the conditions are not at STP? There are two possible solutions. 1. Use the Combined Gas Law to compute the new volume at the new temperature and pressure.

35. Nitric acid is the only acid that will dissolve copper. 20.0 g 4.70 L NO 0.315 mol ??? L 0.210 mol 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O x 0.667

36. What is the volume at a temperature of 25 C, and a pressure of 691 torr? 20.0 g4.70 L (STP) Using the Combined Gas Law 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O

37. Solve for V 2 V 2 = 5.65 L V 2 is the new volume at 25 C and 691 torr

38. Using the Ideal Gas Equation

39. What if the conditions are not at STP? There are two possible solutions. 2. Use the Ideal Gas Equation to compute the new volume using the new temperature and pressure.

40. Moles of gasPressure Volume in Liters The gas constant Temperature in K The value of R depends on pressure units.

41. 20.0 g 4.70 L NO 0.315 mol ??? L 0.210 mol 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O x 0.667 We know that 0.210 mol of NO gas at STP was produced from 20.0g Cu.

42. What is the volume at a temperature of 25 C, and a pressure of 691 torr? 20.0 g0.210 mol Using the Ideal Gas Eauation PV = nRT 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O

43. Solve PV=nRT for V Then plug in the new temperature and pressure. V = 5.65 L

44. Either method works fine in this case, since you have a gas that you know both the pressure, and the number of moles.

45. Use the Combined Gas Law when you are looking for a new pressure, volume or temperature for a confined gas. (moles are constant) Use the Ideal Gas Equation when you are looking for a pressure, volume or temperature when the number of moles is known.

46. Of course the Ideal Gas Equation can be used whenever you are dealing with pressure, volume, temperature or the number of moles, and any one of the variables could be the unknown. Practice solving PV=nRT for each variable.

47. Remember, when using PV = nRT … Pressure is usually in atmospheres, torr, mm Hg, or kilopascals. Volume is usually in liters. … and n is in moles. The value of R, the gas constant, depends on the units of V and P. Temperature must be in Kelvin…

48. Simple ideal gas problem: How many moles are in a 2.00 L container of oxygen gas at a pressure of 0.950 atm and a temperature of 20.0 C?

49. This could actually be part of a larger problem: Find the mass of iron(III) oxide formed when excess iron reacts with 2.00 L of oxygen gas at 0.950 atm and a temperature of 20.0 C.

50. Start with the balanced equation: 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) Excess0.950 atm, 2.00 L, 20.C Use the ideal gas law to find the moles of O 2. PV = nRT = 0.0790 mol

51. Start with the balanced equation: 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) Excess0.950 atm, 2.00 L, 20.C 0.0790 mol0.0527 mol 8.41 g x 2/3 = 8.41g Fe 2 O 3

52. There is a useful variation on the Ideal Gas Equation when dealing with molar mass. Since … Then PV=nRT becomes

53. An example of using

54. Find the volume of 0.325 grams of nitrogen gas at a pressure of 1.20 atm and a temperature of 300. K. Use this equation because you are given the mass of a gas for which you can easily calculate the molar mass.

55. Find the volume of 0.325 grams of nitrogen gas at a pressure of 1.20 atm and a temperature of 300. K. V = 0.238 L Solve for V

56. Look forward to more stoichiometry problems using the Ideal Gas Equation.