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Gases II

Gases II. Dr. Ron Rusay. http://chemconnections.org/general/chem106/Magnesium-Zinc-wo.1.mov. Experimentally Determining The Moles & Molar Mass of Hydrogen Applying Avogadro’s Law. Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g). Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g).

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Gases II

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  1. Gases II Dr. Ron Rusay

  2. http://chemconnections.org/general/chem106/Magnesium-Zinc-wo.1.movhttp://chemconnections.org/general/chem106/Magnesium-Zinc-wo.1.mov Experimentally Determining The Moles & Molar Mass of HydrogenApplying Avogadro’s Law Mg(s) + 2HCl(aq)  MgCl2(aq) + H2 (g) Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2 (g) Using the Ideal Gas Law & Partial Pressures Dr. Ron Rusay

  3. What is wrong with this set up? Mg or Zn

  4. Mg(s) + 2HCl(aq)  MgCl2(aq) + H2 (g) Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2 (g)

  5. Ideal Gas Law PV = n RT • R = “proportionality” constant = 0.08206 L atm  mol • P = pressure in atm • V = volume in liters • n = moles • T = temperature in Kelvins

  6. Standard ConditionsTemperature, Pressure & Moles • “STP” • For 1 mole of a gas at STP: • P = 1 atmosphere • T = C (273.15 K) • The molar volume of an ideal gas is 22.42 liters at STP

  7. Isobaric process: pressure constant Isochoric process: volume constant Isothermal process: temperature constant P1 V1 = P2 V2 P V V1 / n1 = V2 / n2 V n V1 / T1 = V2 / T2 V T

  8. Applying a Gas Law Provide 3 different sets of conditions (Pressure and Temperature) which can account for the volume of the gas decreasing by ½.

  9. Applications If the molecules of a trapped gas maintain the same average velocity when their volume is changed, the pressure will be inversely related to the volume change. Spore Dispersal / Boyle’s Law “Spud Guns” and a Voyage to Near Outer Space

  10. Hydrogen & the Ideal Gas Law n H2(g) = PV/ RT • n = moles H2(g) • P H2(g) = pressure of H2(g) in atm (mm Hg atm) • V = experimental volume (mL L) • T = experimental temperature (oC  K) Mg(s) + 2HCl(aq)  MgCl2(aq) + H2 (g) Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2 (g)

  11. Total Pressure:Sum of the Partial Pressures • For a mixture of gases, the total pressure is the sum of the pressures of each gas in the mixture. PTotal = P1 + P2 + P3 + . . . PTotal n Total nTotal = n1 + n2 + n3 + . . .

  12. P H2(g) = P Total (barometric)- P H2O (g) [TABLE] - P HCl (g) P HCl (g) = HCl Height (mm) x 0.0772__________ Density Hg is 12.95 times > density HCl(aq) P HCl (g) = HCl Height (mm) ÷ 12.95__________ Density Hg is 12.95 times > density HCl(aq) 0.772 mm Hg/cm of acid solution

  13. Ideal Gas Law: Moles & Avogadro’s Law n H2(g) = PV/ RT • n = moles H2(g) • P H2(g) = pressure of H2(g) in atm (mm Hg atm) • P H2(g) = P Total (barometric)- P H2O (g)[TABLE] - P HCl (g) • V = experimental volume (mL L) • T = experimental temperature (oC  K) Mg(s) + 2HCl(aq)  MgCl2(aq) + H2 (g) Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2 (g)

  14. Do you have enough oxygen to climb Mt. Everest? http://chemconnections.org/chemwiki/everest/everest.htm

  15. QUESTION • A typical total capacity for human lungs is approximately 5,800 mL. At a temperature of 37°C (average body temperature) and pressure of 0.98 atm, how many moles of air do we carry inside our lungs when inflated? (R = 0.08206 L atm/ K mol) • A) 1.9 mol • 0.22 mol • 230 mol • 2.20 mol • 0 mol: Moles can harm a person’s lungs.

  16. ANSWER B) is correct. The units for temperature must be in K, pressure in atm, and volume in L. Then using the universal constant 0.08206 L atm/ K mol in the PV = nRT equation moles may be solved. n O2(g) = PV/ RT

  17. An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm a) How many molesof O2 are actually in a typical breath?. b) What is the mass of O2 in a typical breath?. c) How much of the O2 is essential biochemically?

  18. QUESTION • An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm: • How many molesof air are actually in a typical breath? • A) 0.0020 mol • 0.020 mol • 0.030 mol • 0.025 mol • 0.0042 mol

  19. An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm a) How many molesof O2 are actually in a typical breath?. b) What is the mass of O2 in a typical breath?. c) How much of the O2 is essential biochemically? n O2(g) = (20.9%) * PV/ RT nO2(g) = (0.209 molO2(g) / mol air) (1.0 atm x (3.5 L-3.0 L) x mol air * K / 0.0821 L * atm x 300 K) nO2(g)= 0.0042 mol (E)

  20. ANSWER • An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm: • How many molesof air are actually in a typical breath? • A) 0.0020 mol • 0.020 mol • 0.030 mol • 0.025 mol • 0.0042 mol

  21. QUESTION An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm: What is the mass of O2 in a typical breath? A) 0.0042 mol x 16 g/mol B) 0.020 mol x 16 g/mol C) 0.0042 mol x 32 g/mol D) 0.020 mol x 32 g/mol

  22. An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm a) How many molesof O2 are actually in a typical breath?. b) What is the mass of O2 in a typical breath?. c) How much of the O2 is essential biochemically? n O2(g) = (20.9%) * PV/ RT nO2(g) = (0.209 molO2(g) / mol air) (1.0 atm x (3.5 L-3.0 L) x mol air * K / 0.0821 L * atm x 300 K) nO2(g)= 0.0042 mol (E) gO2(g)= 0.0042 mol x 32.0 g/mol gO2(g)= 0.13 g

  23. An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm c) How much of the O2 is essential biochemically? Two estimates for a person with normal physical activity range from 0.67 - 0.84 kg of O2 being used per day (NASA provided the higher value). How many breaths do you take in one day? ~ 5 mol % of the O2is actually used per breath. Hard exercise increases this oxygen demand (intake) about 10 fold.

  24. QUESTION The primary source of exhaled CO2 is from the combustion of glucose, C6H12O6 (molar mass = 180. g/mol.). The balanced equation is shown here: C6H12O6 (aq) + 6 O2 (g)  6 CO2 (g) + 6 H2O (l) If you oxidized 5.42 grams of C6H12O6 while tying your boots to climb Mt. Everest, how many liters of O2 @ STP conditions did you use? A) 0.737 L B) 0.672 L C) 4.05 L D) 22.4 L

  25. ANSWER C), assuming you were not holding your breath, is correct. The number of moles of glucose must first be determined (5.42/180. = 0.0301 moles), then this is multiplied by 6 to account for the stoichiometric ratio between glucose and oxygen. From this, PV = nRT may be used with the appropriate substitutions.

  26. Molar Mass of a Gas (Hydrogen) • PV = nRT • n= g of gas/ MM gas [MM gas = g/mol] • PV = (g of gas/ MM gas)RT • MM gas = g of gas/V (RT/P) Density of gas [separate experiment] • MM gas = g of gas/V (RT/P) • MM gas = density of gas (RT/P)

  27. QUESTION • Freon-12, CF2Cl2, a “safe” compressible gas, was widely used from 1935-1994 as a refrigerant in refrigerators, freezers, and air conditioning systems. However, it had been shown to be a greenhouse gas and to catalytically destroy the ozone layer. It was phased out and banned. • 200 ml of Freon-12 was collected by syringe. It weighed 0.927 grams, had a temperature of 30.0°C, and a pressure of 730 mm of Hg. What is the experimental molar mass of Freon-12? • 12.1 g/mol • 84 g/mol • 92.7 g/mol • 115 g/mol • 121 g/mol

  28. ANSWER • E) is the molar mass of Freon-12. The pressure must be corrected for the presence of water by subtracting 31.8 mmHg from the total pressure. This should also be converted to atm. The temperature must be converted to K. Then PV = nRT can be used if n is written as g/mm and solved for mm. ……. Or use the Periodic Table.

  29. QUESTION • The density of an unknown atmospheric gas pollutant was experimentally determined to be 1.964 g/ L @ 0 oC and 760 torr. • What is the molar mass of the gas? • What might the gas be? A) CO B) SO2 C) H2O D) CO2

  30. ANSWER MM gas = density of gas (RT/P) MM gas = 1.964 g/ L x 0.08206 L atm  mol x 273K/ 760 torr x 760 torr/ 1atm 1.964 g/ L @ 0 oC and 760 torr. R = 0.08206 L atm  mol oC  K torr  atm MM gas = 44.0 g/mol MM gas = D) CO2

  31. QUESTION 0.0820 grams of the volatile, gaseous phase, of a compound, which smells like fresh raspberries, was trapped in a syringe. It had a volume of 12.2 mL at 1.00 atmosphere of pressure and 25.0°C. What is the molar mass of this pleasant smelling compound ? A) 13.8 g/mol B) 164 g/mol C) 40.9 g/mol D) 224 g/mol

  32. ANSWER B) 164 g/mol Using PV = nRT : 0.0122 L for V, 298 K for T, 0.08206 for R and solving for n the number of moles represented by 0.0820 grams can be obtained. Then the MOLAR MASS (grams in one mole) can be determined. MM gas = density of gas (RT/P) MM gas = 0.0820 g/ L x 0.00122 L x 0.0821 atm  mol x 298K/ 1atm

  33. QUESTION For the compound that smells like fresh raspberries, the following structure matches its molecular formula. A) TRUE B) FALSE

  34. ANSWER Based on your answers for the compound, which smells like fresh raspberries, in the previous two questions, the following structure matches its molecular formula. A) TRUE B) FALSE 164 g/mol = C10H12O2

  35. QUESTION Which sequence represents the gases in order of increasing density at STP? A) Fluorine < Carbon monoxide < Chlorine < Argon B) Carbon monoxide < Fluorine < Argon < Chlorine C) Argon < Carbon monoxide < Chlorine < Fluorine D) Fluorine < Chlorine < Carbon monoxide < Argon

  36. ANSWER Which sequence represents the gases in order of increasing density at STP? A) Fluorine < Carbon monoxide < Chlorine < Argon B) Carbon monoxide < Fluorine < Argon < Chlorine C) Argon < Carbon monoxide < Chlorine < Fluorine D) Fluorine < Chlorine < Carbon monoxide < Argon

  37. Gases & AirbagsUse of Chemical Reactions and Physical Properties

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