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Basic Stoichiometry

Basic Stoichiometry. Pisgah High School M. Jones Revision history: 5/16/03, 02/04/12, 04/27/12. With the Gas Laws. The word stoichiometry comes from the Greek words stoicheion which means “element” and metron which means “measure”.

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Basic Stoichiometry

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  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 = 6.022 x1023 atoms or molecules 1 mole = the molar mass 1 mole = 22.4 L of any gas at STP

  6. The word mole is one that represents a very large number. Much like “dozen” means 12, … “mole” means 6.022 x 1023

  7. The key to doing stoichiometry is the balanced chemical equation. 2 H2 + O2 2 H2O 2 2

  8. The coefficients give the relative number of atoms or molecules of each reactant or product … as well as the number of moles. 2 H2 + O2 2 H2O

  9. 2 H2 + O2 2 H2O 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 H2 + O2 2 H2O 2 molecules of hydrogen 1 molecule of oxygen 2 molecules of water The balanced chemical equation also gives the smallest integer number of moles.

  11. 2 H2 + O2 2 H2O 2 moles of hydrogen 1 mole of oxygen 2 moles of water The balanced chemical equation also gives the smallest integer number of moles.

  12. 2 H2 + O2 2 H2O 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 H2 + O2 2 H2O 2 moles 1 mole 2 moles 2 L 1 L 2 L

  15. C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) 5 mols 3 mols 4 mols 1 mol 5 L 3 L 4 L 1 L 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 O2

  16. C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) 5 mols 3 mols 4 mols 1 mol 5 L 3 L 4 L 1 L 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 O2 How many liters of carbon dioxide gas and water vapor at STP would be produced? 10.5 L CO2 and 14.0 L H2O

  17. CH4(g) + 4 Cl2(g)  CCl4(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 Cl2

  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. Cl2(g) + H2O2 (l)  2 HOCl (l) 2.80 L ??? g Convert 2.80 L of Cl2 gas at STP to moles

  20. Cl2(g) + H2O2 (l)  2 HOCl (l) 2.80 L ??? g .125 x 2 0.250 mol 0.125 mol 0.125 mol Cl2 13.1g HOCl

  21. The reaction between copper and nitric acid

  22. Will copper dissolve in acids? Consider hydrochloric acid Cu + 2HCl  CuCl2 + 2H2 (g) No Reaction Most metals react with HCl to produce a metal chloride solution and H2 gas. Not copper

  23. Will copper dissolve in acids? Consider hydrochloric acid Cu + 2HCl  CuCl2 + 2H2 (g) No Reaction 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? Cu + HBr  NR Cu + HI  NR Cu + HF  NR Cu + H2SO4  NR Cu + HC2H3O2  NR The same is true for all acids except nitric acid

  25. A beaker contains a penny and some nitric acid is added. Nitric acid is the only acid that will dissolve copper. 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

  26. Nitric acid is the only acid that will dissolve copper. 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O The penny begins to disappear and the solution turns blue-green and a brown gas is given off.

  27. Nitric acid is the only acid that will dissolve copper. 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O The gas produced in the reaction is NO, which is colorless. Reddish brown NO2 forms when NO reacts with the oxygen in the air.

  28. Nitric acid is the only acid that will dissolve copper. 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O The penny is gone and the solution turns a dark blue. The brown NO2 gas escapes from the open beaker.

  29. Nitric acid is the only acid that will dissolve copper. 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O Calculate the volume of NOgas at STP when 20.0 grams of copper dissolves.

  30. Nitric acid is the only acid that will dissolve copper. 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O 20.0 g ??? L x 0.667 0.210 mol 0.315 mol 0.315 mol Cu

  31. Nitric acid is the only acid that will dissolve copper. 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O 20.0 g ??? L x 0.667 0.210 mol 0.315 mol 4.70 L NO

  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. 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O 20.0 g ??? L x 0.667 0.210 mol 0.315 mol 4.70 L NO

  36. What is the volume at a temperature of 25 C, and a pressure of 691 torr? 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O 20.0 g 4.70 L (STP) Using the Combined Gas Law

  37. for V2 Solve V2 = 5.65 L V2 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. Pressure Volume in Liters Temperature in K Moles of gas The gas constant The value of R depends on pressure units.

  41. We know that 0.210 mol of NO gas at STP was produced from 20.0g Cu. 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O 20.0 g ??? L x 0.667 0.210 mol 0.315 mol 4.70 L NO

  42. What is the volume at a temperature of 25 C, and a pressure of 691 torr? 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O 20.0 g 0.210 mol Using the Ideal Gas Eauation PV = nRT

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

  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) + 3O2(g)  2Fe2O3(s) Excess 0.950 atm, 2.00 L, 20.C Use the ideal gas law to find the moles of O2. PV = nRT = 0.0790 mol

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