1. Gases and gas laws
Chapters 13.1 and 14

2. Kinetic molecular theory The theory of moving molecules
Basic assumptions describing the behavior of gases:
Gas consist of small particles that are very far apart
Most of the volume of a gas is empty space
Individual particles have essentially zero volume compared to the total volume of the gas
Gases particles are in constant, random, straight line motion
Attractive and repulsive forces between particles are negligible

3. Kinetic molecular theory The theory of moving molecules
Basic assumptions:
Random motion of gas particles is constantly interrupted by collisions between particles and between particles and surfaces
Such collisions are perfectly elastic
Kinetic energy may be transferred between particles but is not lost (conserved)
The average kinetic energy of gas particles is directly proportional to absolute temperature (kelvin)
The rate of motion of particles is related to temperature

4. Kinetic molecular theory The theory of moving molecules
Characteristics of gases:
Gases readily change volume
Container containing gas is always full
Molecules separate and spread out uniformly until fill available space
Compressibility: can be compressed or reduced to small fractions of original volume
Expandability: can expand to fill virtually any volume
Gases typically have low densities (mass to volume ratio)

5. Kinetic molecular theory The theory of moving molecules
Characteristics of gases:
Gases exert pressure
Due to constant random motion, molecules collide with each other and with the walls of their container
Collisions exert pressure on surroundings

6. Kinetic molecular theory The theory of moving molecules
Characteristics of gases:
Gas Temperature and Kinetic Energy
Relationship between average kinetic energy of group of gas molecules and the temperature measured in degrees Kelvin is directly proportional
As temperature increases, average kinetic energy of molecules ____________
As temperature decreases, average kinetic energy of molecules ____________
Relationship only exists when temperature is expressed in Kelvin scale
Molecules have 0 kinetic energy when temp is 0 K
Only have kinetic energy at temps above 0 K
increases
decreases

7. Movement of gas molecules
Describes movement of gases through other materials
Gas molecules move from areas of high concentration to low concentration
Examples:
Perfume molecules spreading across a room
Aromas of cooking foods waft through the house
DIFFUSION
Animation by Caitlin Claunch

8. Movement of gas molecules
Gas molecules escape through tiny openings
Gas molecules move from an area of high pressure to low pressure
Examples:
Tire flattens due to a puncture
Balloon deflates when stuck with a pin
EFFUSION
Animation by Caitlin Claunch

9. Gases Review
Gas particles are in constant, ________ motion. Particles travel in ____________.
The size of the particles is ___________ compared to the volume of ______ space.
Attraction and repulsion between particles is __________.
Collisions are ______ because particles transfer energy but the total kinetic energy is ________.
____________ is the measure of average kinetic energy of particles. At a given temperature, ___ gas particles have the same ______________ energy.
In diffusion, gas particles move from high ____________ to low ____________, whereas in effusion, particles move from high ________ to low ________.
random
straight lines
insignificant
empty
negligible
elastic
constant
Temperature
all
average kinetic
concentration
concentration
pressure
pressure

10. Variables describing gases
Pressure (P): force per unit area
Volume (V): amount of space taken up
Temperature (T): measure of average kinetic energy
Number of Moles (n): amount of substance

11. pressure
Results from billions of gas particles in motion and colliding against their container
Barometer: instrument used to measure atmospheric pressure
Manometer: instrument used
to measure gas pressure in a
closed container
Worksheet: Pressure Unit Conversions

12. pressure
UNITS:
The relationships among the different units of pressure can be used as conversion factors
Values expressed in table above describe the same amount of pressure
atm
mm Hg
torr
psi
kPa Pa 1
760
14.7
101,325
SI UNIT
Worksheet: Pressure Unit Conversions

13. Dalton’s law of partial pressures
The total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture
Ptotal = P1 + P2 + P3 + … Pn
101.3
kPa
kPa kPa kPa kPa kPa

14. Finding the partial pressure
A mixture of oxygen (O2), carbon dioxide (CO2), and nitrogen (N2) has a total pressure of atm. What is the partial pressure of O2, if the partial pressure of CO2 is 0.70 atm and the partial pressure of N2 is 0.12 atm?
Known
PN2 = 0.12 atm
PCO2 = 0.70 atm
Ptotal = 0.97 atm
Unknown
PO2 = ? atm
Practice Problems #4, #5, #6 – p. 392
Ptotal = PO2 + PCO2 + PN2
PO2 = Ptotal – PCO2 – PN2
PO2 = 0.97 atm – 0.70 atm – 0.12 atm =
0.15 atm

15. U12-3
PRACTICE/EXAMPLES
p. 422: #1, #2, #3, #4, #5; p. 425: #6, #7, #8; p. 427: #9, #10, #11, #12, #13

16. U12-3: p. 392, #4
What is the partial pressure of hydrogen gas in a mixture of hydrogen and helium if the total pressure is 600 mm Hg and the partial pressure of helium is 439 mm Hg?
Known
PHe = 439 mm Hg
Ptotal = 600 mm Hg
Unknown
PH2 = ? mm Hg
U12-3 Practice Examples: #4, #5, #6 – p. 392
Ptotal = PH2 + PHe
PH2 = Ptotal – PHe
PH2 = 600 mm Hg – 439 mm Hg =
161 mm Hg

17. U12-3: p. 392, #5
Find the total pressure for a mixture that contains four gases with partial pressures of 5.00 kPa, 4.56 kPa, 3.02 kPa, and 1.20 kPa.
Known
P1 = 5.00 kPa
P2 = 4.56 kPa
P3 = 3.02 kPa
P4 = 1.20 kPa
Unknown
Ptotal = ? kPa
U12-3 Practice Examples: #4, #5, #6 – p. 392
Ptotal = P1 + P2 + P3 + P4
Ptotal = 5.00 kPa kPa kPa kPa
= kPa

18. U12-3: p. 392, #6
Find the partial pressure of carbon dioxide in a gas mixture with a total pressure of 30.4 kPa if the partial pressures of the other two gases in the mixture are 16.5 kPa and 3.7 kPa.
Known
P1 = 16.5 kPa
P2 = 3.7 kPa
Ptotal = 30.4 kPa
Unknown
PCO2 = ? kPa
U12-3 Practice Examples: #4, #5, #6 – p. 392
Ptotal = P1 + P2 + PCO2
PCO2 = Ptotal – P1 – P2
PCO2 = 30.4 kPa – 16.5 kPa – 3.7 kPa
= 10.2 kPa

19. p. 415: #68, #69, #70, #71
Chapter 13 homework

20. Pressure Unit Conversions Worksheet

21. volume Gases expand to fill their container
That is, they do not have a definite volume
Gas laws explain how and when gas volume changes
UNITS: L mL
cm3
quart 1
1000
1.057
SI UNIT

22. temperature
Measure of average kinetic energy of particles
Recall KMT: At any given temperature, all gases have same average kinetic energy
UNITS:
Measured using the Kelvin scale (absolute temperature scale) developed by Lord Kelvin
0 degrees kelvin is defined as absolute zero
Absolute zero is the hypothetical temperature at which all motion stops
K = °C + 273

23. Stp Standard Temperature and Pressure 0°C or 273 K
1 atm or kPa

24. Combined gas law
States the relationship among pressure (P), volume (V), and temperature (T) when amount of gas (n) remains constant
Pressure and Temperature are
_______________
proportional
directly
Pressure and Volume are
_______________
proportional
inversely
Volume and Temperature are
_______________
proportional
directly

25. Using Combined gas law
A gas occupies 7.84 cm3 at 71.8 kPa and 25°C. What volume will it occupy at STP?
Rearrange to solve for unknown variable, V2
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
71.8 kPa
7.84 cm3
25°C
+ 273 K = 298 K
kPa
unknown
273 K

26. Using Combined gas law V2 = (71.8 kPa)(7.84 cm3)(273K)
A gas occupies 7.84 cm3 at 71.8 kPa and 25°C. What volume will it occupy at STP?
V2 = (71.8 kPa)(7.84 cm3)(273K)
( kPa)(298K)
V2 = 5.1 cm3
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
71.8 kPa
7.84 cm3
25°C
+ 273 K = 298 K
kPa
unknown
273 K
p #19, #20, #21, #22, #23

27. p. 449: #92, #93, #94
Chapter 14 homework

28. U12-3
PRACTICE/EXAMPLES
p. 422: #1, #2, #3, #4, #5; p. 425: #6, #7, #8; p. 427: #9, #10, #11, #12, #13

29. U12-3: p. 430, #19
A helium-filled balloon at sea level has a volume of 2.1 L at atm and 36°C. If it is released and rises to an elevation at which the pressure is atm and the temperature is 28°C, what will be the new volume of the balloon?
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
0.998 atm
2.1 L
36°C
+ 273 K = 303 K
0.900 atm
V2 = 2.3 L
unknown
28°C
+ 273 K = 301 K

30. U12-3: p. 430, #20
At 0.00°C and 1.00 atm pressure, a sample of gas occupies 30.0 mL. If the temperature is increased to 30.0°C and the entire gas sample is transferred to a 20.0-mL container, what will be the gas pressure inside the container?
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
1.00 atm
P2 =
P1V1T2
V2T1
30.0 mL
0.00°C
+ 273 K = 273 K
unknown
P2 = 2.3 L
20.0 mL
30.0°C
+ 273 K = 303 K

31. Boyle’s law Boyle’s Law
States the relationship between pressure (P) and volume (V), when temperature (T) and amount of substance (n) remain constant
Boyle’s Law
Pressure and Volume are __________ proportional
As pressure increases, volume __________
As pressure decreases, volume __________
inversely
decreases
increases

32. Using boyle’s law
A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
Rearrange to solve for unknown variable, V2
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
150. kPa
100. mL
remains constant
200. kPa
unknown
remains constant

33. Using boyle’s law
A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
150. kPa
100. mL
remains constant
200. kPa
unknown
V2 = (150. kPa)(100. mL)
(200. kPa)
V2 = mL
p #1, #2, #3, #4, #5

34. Charles’ law Charles’ Law
States the relationship between volume (V) and temperature (T), when pressure (P) and amount of substance (n) remain constant
Charles’ Law
Volume and Temperature are __________ proportional
As volume increases, temperature __________
As volume decreases, temperature __________
directly
increases
decreases

35. Using charles’ law
A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.
Rearrange to solve for unknown variable, V2
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
remains constant
473 cm3
36°C
+ 273 K = 309 K
remains constant
unknown
94°C
+ 273 K = 367 K

36. Using charles’ law
A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
V2 = (473 cm3)(367 K)
(309 K)
V2 = 560 cm3
remains constant
473 cm3
p #6, #7, #8
36°C
+ 273 K = 309 K
remains constant
unknown
94°C
+ 273 K = 367 K

37. Gay-lussac’s law Gay-Lussac’s Law
States the relationship between pressure (P) and temperature (T), when volume (V) and amount of substance (n) remain constant
Gay-Lussac’s Law
Pressure and Temperature are __________ proportional
As pressure increases, temperature __________
As temperature decreases, pressure __________
directly
increases
decreases

38. Using gay-lussac’s law
The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C, what will be the gas pressure in the tank?
Rearrange to solve for unknown variable, P2
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
3.20 atm
remains constant
22.0°C
+ 273 K = 295 K
unknown
remains constant
60.0°C
+ 273 K = 333 K

39. Using gay-lussac’s law
The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C, what will be the gas pressure in the tank?
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
3.20 atm
remains constant
unknown
22.0°C
+ 273 K = 295 K
60.0°C
+ 273 K = 333 K
P2 = (3.20 atm)(333.0K)
(295.0K)
P2 = atm
p #9, #10, #11, #12, #13

40. p. 448: #88, #89, #90, #91
U12-5: Ch.14 HOMEWORK

41. U12-3
PRACTICE/EXAMPLES
p. 422: #1, #2, #3, #4, #5; p. 425: #6, #7, #8; p. 427: #9, #10, #11, #12, #13

42. U12-3: p. 430, #21
A sample of air in a syringe exerts a pressure of atm at a temperature of 22.0°C. The syringe is placed in a boiling water bath at 100.0°C. The pressure of the air is increased to 1.23 atm by pushing the plunger in, which reduces the volume to mL. What was the original volume of air?
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =

43. U12-3: p. 422, #1
The volume of a gas at 99.0 kPa is mL. If the pressure is increased to 188 kPa, what will be the new volume?
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =

44. U12-3: p. 425, #6
A gas at 89°C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L?
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =

45. U12-3: p. 427, #9
A gas in a sealed container has a pressure of 125 kPa at a temperature of 30.0°C. If the pressure in the container is increased to 201 kPa, what is the new temperature?
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =

46. U12-3: p. 422, #2
The pressure of a sample of helium in a 1.00-L container is atm. What is the new pressure if the sample is placed in a 2.00-L container?
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =

47. U12-3: p. 430, #22
An unopened, cold 2.00-L bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm at a temperature of 5.0°C. If the bottle is dropped into a lake and sinks to a depth at which the pressure is 2.85 atm and the temperature is 2.09°C, what will be the volume of gas in the bottle?
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =

48. U12-3: p. 422, #3
Air trapped in a cylinder fitted with a piston occupies mL at 1.08 atm pressure. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston?
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =

49. U12-3: p. 430, #23
A sample of gas of unknown pressure occupies L at a temperature of 298 K. The same sample of gas is then tested under known conditions and has a pressure of 32.6 kPa and occupies L at 303 K. What was the original pressure of the gas?
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =

50. U12-3: p. 427, #12
If a gas sample has a pressure of 30.7 kPa at 0.00°C, by how much does the temperature have to decrease to lower the pressure to 28.4 kPa?
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =

51. Avogadro’s Law Conversion factor 22.4 L of any gas = 1 mol
Addresses variable n (number of moles)
States that equal volumes of gases at the same temperature and pressure contain the same numbers of particles
Molar volume of gases: 22.4 L/mol at standard temperature and pressure (STP)
Standard temperature: 273 K or 0°C
Standard pressure: 1.00 atm, kPa, 760 torr, 760 mm Hg
Conversion factor
22.4 L of any gas = 1 mol

52. Using avogadro’s law
Calculate the volume that mol of gas at STP will occupy. Is amount of substance (n) involved? Is gas at STP? Use relationship between volume and number of moles (Avogadro’s Law)
Yes
Yes

53. Using avogadro’s law
Calculate the volume that mol of gas at STP will occupy. Conversion factor: 22.4 L = 1 mol Unit for volume?
Liter (L)
L = mol

54. Using avogadro’s law
Calculate the volume that mol of gas at STP will occupy. Conversion factor: 22.4 L = 1 mol
L = mol mol

55. Using avogadro’s law
Calculate the volume that mol of gas at STP will occupy. Conversion factor: 22.4 L = 1 mol
L = mol mol

56. Using avogadro’s law
Calculate the volume that mol of gas at STP will occupy. Conversion factor: 22.4 L = 1 mol
L = mol L mol

57. Using avogadro’s law
Calculate the volume that mol of gas at STP will occupy. Conversion factor: 22.4 L = 1 mol
L = mol 22.4 L 1 mol

58. Using avogadro’s law
Calculate the volume that mol of gas at STP will occupy.
L = mol 22.4 L 1 mol L = 19.7 L

59. Using avogadro’s law
Calculate the volume that 2.0 kg of methane gas (CH4) will occupy at STP.
L = 2.0 kg L = 2800 L
1000 g
1 mol
22.4 L
1 kg
16.0 g
1 mol
P #29, #30, #31, #32, #33

60. U12-2: Ch. 13 homework U12-5: Ch. 14 homework p. 449: #95, #96
Express answers in atm, kPa, and torr.
U12-2: Ch. 13 homework
p. 449: #95, #96
U12-5: Ch. 14 homework

61. U12-3
PRACTICE/EXAMPLES
p. 432: #24, #25, #26, #27, #28; p. 433: #29, #30, #31, #32, #33

62. U12-3: p. 432, #24
Determine the volume of a container that holds 2.4 mol of gas at STP.
U12-3: p. 432, #25
What size container do you need to hold mol N2 gas at STP?

63. U12-3: p. 432, #26
What volume will 1.02 mol of carbon monoxide gas occupy at STP?
U12-3: p. 432, #27
How many moles of nitrogen gas will be contained in a 2.00-L flask at STP?

64. U12-3: p. 432, #28
If a balloon will rise off the ground when it contains mole of helium in a volume of L, how many moles of helium are needed to make the balloon rise when its volume is L? Assume that temperature and pressure stay constant.

65. U12-3: p. 433, #29
How many grams of carbon dioxide gas are in a 1.0-L balloon at STP?
U12-3: p. 433, #30
What volume in milliliters will g H2 gas occupy at STP?

66. Ideal gas law
Describes the physical behavior of an ideal gas in terms of all four variables
Pressure (P)
Volume (V)
Temperature (T)
Amount of gas (n)

67. Ideal gases vs. Real gases
Described by Kinetic Molecular Theory
Particles have no volume, no intermolecular attractive forces, and perfectly elastic collisions
Follow all gas laws under all conditions of temperature and pressure
Behave less like ideal gases at extremely high pressures and low temperatures
Polar gases do not behave as ideal gases
Larger gas molecules behave less like ideal gases than smaller gas molecules
IDEAL GASES
REAL GASES

68. Ideal gas law Combines: Boyle’s Law Charles’ Law Gay-Lussac’s Law
Avogadro’s Law
Combined Gas Law
If number of moles (n) is constant → Combined Gas Law
If number of moles (n) is not constant, and
At STP → Avogadro’s Law
Not at STP → Ideal Gas Law

69. How will you know which value to use?
Ideal gas constant (r)
Experimentally determined
Three possible numerical values
How will you know which value to use?

70. Using the ideal gas law
A sample of nitrogen gas (N2) has a volume of liters at 0.00°C and 1.50 atm pressure. How many moles of nitrogen are present?
Rearrange to solve for unknown variable, n
VARIABLES
P = atm
V = L
n = ? mol
R =
T = 273 K

71. Using the ideal gas law
A sample of nitrogen gas (N2) has a volume of liters at 0.00°C and 1.50 atm pressure. How many moles of nitrogen are present?
VARIABLES
P = atm
V = L
n = ? mol
R =
T = 273 K
n = (1.50 atm)(5.56 L)
( ( )(273 K)
n = mol

72. U12-3
PRACTICE/EXAMPLES
p. 432: #24, #25, #26, #27, #28; p. 433: #29, #30, #31, #32, #33

73. Which gas law? Avogadro’s Law 22.4 L = 1 mol
What volume does 9.45 g of C2H2 occupy at STP?
L = 9.45 g C2H2 L = 8.14 L
1 mol C2H2
22.4 L C2H2
26.0 g C2H2
1 mol C2H2
P #29, #30, #31, #32, #33

74. Which gas law? Ideal Gas Law PV = nRT
What volume does 9.45 g of C2H2 occupy at STP?
Rearrange to solve for unknown variable, V
VARIABLES
P = kPa
V = ? L
n = 9.45 g = mol
26.0 g/mol
R =
T = 273 K

75. Which gas law? What volume does 9.45 g of C2H2 occupy at STP?
VARIABLES
P = kPa
V = ? L
n = 9.45 g = mol
26.0 g/mol
R =
T = 273 K

76. Which gas law? V = (0.3634615385 mol)( )(273 K) (101.325 kPa)
What volume does 9.45 g of C2H2 occupy at STP?
V = ( mol)( )(273 K)
( kPa)
V = 8.14 L

77. Which gas law?
You decide to go on a long hot air balloon ride, and pack a bottle of shampoo for your trip. There is some gas inside the shampoo bottle when you climb into the basket at the beginning of your journey. Being the good scientist that you are, you decide to take constant measurements of your surroundings. The shampoo bottle contains 435 mL of gas, is under a pressure of 1.10 atm, and is at a temperature of 30.0°C. As you climb high into the air, the bottle starts to expand and eventually explodes, covering you and your companions with Pert Plus®. Eager to explain this phenomenon, you take some quick measurements, noting that the pressure has dropped to atm and the temperature is now 5.00°C. To what new volume did the gas inside the shampoo bottle expand?
VARIABLES
P1 =
V1 =
T1 =
1.10 atm
P2 =
V2 =
T2 =
0.734 atm
435 mL
? = 598 mL ?
30.0°C = 303 K
5.00°C = 278 K

78. U12-1
Gas Laws Notes
p. 432: #24, #25, #26, #27, #28; p. 433: #29, #30, #31, #32, #33

79. Gas stoichiometry
Stoichiometric calculations always require a balanced equation
N2(g) H2(g) → NH3(g)
Coefficients indicate ratios of reactants and products
1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas OR
1 volume of nitrogen gas reacts with 3 volumes of hydrogen gas to produce 2 volumes of ammonia gas
Volume ratios only work when all reactants and products are gases
(1) 3 2

80. Gas stoichiometry
N2(g) + 3H2(g) → 2NH3(g) How many liters of gaseous ammonia will be made from 5.00 L of hydrogen gas reacting with excess nitrogen gas?
L NH3 = 5.00 L H2

81. Gas stoichiometry
N2(g) + 3H2(g) → 2NH3(g) How many liters of gaseous ammonia will be made from 5.00 L of hydrogen gas reacting with excess nitrogen gas?
L NH3 = 5.00 L H2
L H2

82. Gas stoichiometry
N2(g) + 3H2(g) → 2NH3(g) How many liters of gaseous ammonia will be made from 5.00 L of hydrogen gas reacting with excess nitrogen gas?
L NH3 = 5.00 L H2 L NH3
L H2
= L NH3 2 3

83. U12-10
Volume-Volume WS

84. U12-1
Gas Laws Notes
p. 432: #24, #25, #26, #27, #28; p. 433: #29, #30, #31, #32, #33

85. Gas stoichiometry
Gas stoichiometry problems can involve mass and volume Zn(s) + HCl(aq) → H2(g) + ZnCl2(aq) What volume (in L) of hydrogen can be produced when 6.54 g of zinc reacts with hydrochloric acid at STP? 2
L H2 = 6.54 g Zn

86. Gas stoichiometry
Gas stoichiometry problems can involve mass and volume Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) What volume (in L) of hydrogen can be produced when 6.54 g of zinc reacts with hydrochloric acid at STP?
L H2 = 6.54 g Zn 1 mol Zn
65.4 g Zn

87. Gas stoichiometry
Gas stoichiometry problems can involve mass and volume Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) What volume (in L) of hydrogen can be produced when 6.54 g of zinc reacts with hydrochloric acid at STP?
L H2 = 6.54 g Zn 1 mol Zn 1 mol H2
65.4 g Zn 1 mol Zn

88. Gas stoichiometry
Gas stoichiometry problems can involve mass and volume Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) What volume (in L) of hydrogen can be produced when 6.54 g of zinc reacts with hydrochloric acid at STP?
L H2 = 6.54 g Zn 1 mol Zn 1 mol H L H2
65.4 g Zn 1 mol Zn 1 mol H2
2.24 L H2

89. Gas stoichiometry
Gas stoichiometric calculations only involving volume do not take the temperature and pressure conditions into consideration
After mixing, each gas is at the same temperature and pressure
Gas stoichiometric calculations involving volume and mass must take temperature and pressure conditions into account
volume-volume relationships from balanced chemical equation are combined with
the Ideal Gas Law

90. Gas stoichiometry
When 3.00 L of propane gas is completely combusted to form water vapor and carbon dioxide at a temperature of 350°C and a pressure of atm, what mass of water vapor will result? (p. 449, #102)
Determine volume of water produced (volume-volume relationships).
C3H8 (g) + 5O2 (g) → 4H2O (g) + 3CO2 (g)
C3H8 (g) + O2 (g) → H2O (g) + CO2 (g)
L H2O = L C3H8
4 L H2O
= 12.0 L H2O
1 L C3H8

91. C3H8 (g) + 5O2 (g) → 4H2O (g) + 3CO2 (g)
Gas stoichiometry
When 3.00 L of propane gas is completely combusted to form water vapor and carbon dioxide at a temperature of 350°C and a pressure of atm, what mass of water vapor will result? (p. 449, #102)
Determine number of moles under given conditions (Ideal Gas Law).
C3H8 (g) + 5O2 (g) → 4H2O (g) + 3CO2 (g)
n = mol
(0.990 atm)(12.0 L H2O)
( )(623 K)
Latm
molK =

92. C3H8 (g) + 5O2 (g) → 4H2O (g) + 3CO2 (g)
Gas stoichiometry
When 3.00 L of propane gas is completely combusted to form water vapor and carbon dioxide at a temperature of 350°C and a pressure of atm, what mass of water vapor will result? (p. 449, #102)
C3H8 (g) + 5O2 (g) → 4H2O (g) + 3CO2 (g)
g H2O = mol H2O
18.0 g H2O
1 mol H2O
g H2O = 4.2 g H2O
g H2O = g H2O

93. U12-10
Gas Stoichiometry WS

94. p #100, #103, #104
U12-5 Ch. 14 homework
g C3H8
103. 2KClO3 (s) → 2KCl (s) + 3O2 (g)
5.70 L O2
104. a. 2CO (g) + 2NO (g) → N2 (g) + 2CO2 (g)
b. 2 volumes CO
2 volumes CO2
c L N2
volume = L, mL or cm3

95. U12-3
PRACTICE/EXAMPLES

96. U12-3: p. 437, #41
If the pressure exerted by a gas at 25°C in a volume of L is 3.81 atm, how many moles of gas are present? Variable(s) involved? Gas law/application? P: V: n: R: T:
P, V, T, n
Ideal Gas Law
3.81 atm
0.044 L
n = mol ?
Latm
molK
0.0821
298 K

97. U12-3: p. 433, #33
Calculate the volume that 4.5 kg of ethylene gas (C2H4) will occupy at STP. Variable(s) involved? Gas law/application?
V, n (mass)
@STP
Avogadro’s Law
3600 L C2H4

98. U12-3: p. 425, #7
The Celsius temperature of a 3.00-L sample of gas is lowered from 80.0°C to 30.0°C. What will be the resulting volume of this gas? Variable(s) involved? Gas law/application?
V, T
Combined Gas Law
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
V2 = L

99. U12-3: p. 441, #57
Determine the volume of hydrogen gas needed to react completely with 5.00 L of oxygen gas to form water.
Variable(s) involved?
Gas law/application?
Write the balanced chemical equation.
2H2 (g) + O2 (g) → 2H2O (g)
Unknown:
Given:
Answer: V
volume-volume stoichiometry
L H2
5.00 L O2
10.0 L H2

100. U12-3: p. 437, #43
Calculate the volume that a mol sample of a gas will occupy at 265 K and a pressure of atm. Variable(s) involved? Gas law/application? P: V: n: R: T:
P, V, T, n
Ideal Gas Law
0.900 atm ?
V = 7.81 L
0.323 mol
Latm
molK
0.0821
265 K

101. U12-3: p. 437, #44
What is the pressure in atmospheres of a mol sample of helium gas at a temperature of 20.0°C if its volume is L? Variable(s) involved? Gas law/application? P: V: n: R: T:
P, V, T, n
Ideal Gas Law ?
0.505 L
P = 5.14 atm
0.108 mol
Latm
molK
0.0821
293 K

102. U12-3: p. 427, #13
A rigid plastic container holds 1.00 L methane gas at 660 torr pressure when the temperature is 22.0°C. How much more pressure will the gas exert if the temperature is raised to 44.6°C? Variable(s) involved? Gas law/application?
P, T
Combined Gas Law
VARIABLES
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
P2 = 711 torr
Additional pressure of 51 torr will be exerted.

103. U12-3: p. 441, #56
What volume of oxygen is needed to react with solid sulfur to form 3.5 L SO2?
Variable(s) involved?
Gas law/application?
Write the balanced chemical equation.
S (s) + O2 (g) → SO2 (g)
Unknown:
Given:
Answer: V
volume-volume stoichiometry
L O2
3.5 L SO2
3.5 L O2