Introduction to Introduction to Gas Laws Gas Laws
Introduction to Gas Introduction to Gas LawsLaws
Introduction to Gas Introduction to Gas LawsLaws
DefinitionsDefinitionsDefinitionsDefinitions
Pressure - the exertion of a force upon a surface by an object in contact with it (force per unit area)
Volume -the amount of 3-dimensional space occupied by an object
Temperature - a measure of the warmth or coldness of an object or substance with reference to another value
Pressure - the exertion of a force upon a surface by an object in contact with it (force per unit area)
Volume -the amount of 3-dimensional space occupied by an object
Temperature - a measure of the warmth or coldness of an object or substance with reference to another value
What are Gas Laws?What are Gas Laws?What are Gas Laws?What are Gas Laws?
Gas Laws study the relationship between these terms
Bed of Nails video
Gas Laws study the relationship between these terms
Bed of Nails video
Bed of Nails VideoBed of Nails VideoBed of Nails VideoBed of Nails Video
QuickTime™ and a decompressor
are needed to see this picture.
PressurePressurePressurePressure
Bed of Nails Video - How does this work?
Newspaper/Ruler Demo - How does this work?
How do does pressure play a role in skating on thin ice? How would you rescue somebody?
Why does pumping air into a car tire increase pressure?
Bed of Nails Video - How does this work?
Newspaper/Ruler Demo - How does this work?
How do does pressure play a role in skating on thin ice? How would you rescue somebody?
Why does pumping air into a car tire increase pressure?
PressurePressurePressurePressurePressure = force/unit area
Bed of Nails video - 1/3 of 8500 lbs is 2833 lbs
2833lbs/3 inches = 944 psi (so at every spike it feels like there is 944 pounds on him)
Pressure = force/unit area
Bed of Nails video - 1/3 of 8500 lbs is 2833 lbs
2833lbs/3 inches = 944 psi (so at every spike it feels like there is 944 pounds on him)
PressurePressurePressurePressure
Pressure is directly proportional to the number of molecules
If you double the mols, you will double the pressure (car tire)
units are atm(atmospheres), mm Hg(millimeters of Mercury), psi(pounds per square inch), Pa(pascals), and torr(torr)
Pressure is directly proportional to the number of molecules
If you double the mols, you will double the pressure (car tire)
units are atm(atmospheres), mm Hg(millimeters of Mercury), psi(pounds per square inch), Pa(pascals), and torr(torr)
VolumeVolumeVolumeVolume
Volume = Length X Width X Height
Volume is inversely proportional to pressure
Double the volume and you will have half the pressure
Units are L, mL, cm3 or m3
Volume = Length X Width X Height
Volume is inversely proportional to pressure
Double the volume and you will have half the pressure
Units are L, mL, cm3 or m3
TemperatureTemperatureTemperatureTemperature
How is temperature different from heat?
Measured in ˚C and converted to Kelvins
Temperature is directly proportional to pressure
Doubling the Temperature will double the pressure
How is temperature different from heat?
Measured in ˚C and converted to Kelvins
Temperature is directly proportional to pressure
Doubling the Temperature will double the pressure
Gas PropertiesGas PropertiesGas PropertiesGas Properties
Gases expand or compress to fill their container, this means that the volume is variable
Gases have very low densities compared to other states of matter
Gases expand or compress to fill their container, this means that the volume is variable
Gases have very low densities compared to other states of matter
Kinetic Molecular TheoryKinetic Molecular TheoryKinetic Molecular TheoryKinetic Molecular Theory
Based on an Ideal Gas
an Ideal Gas has the following properties:
volume of each particle is essentially zero (perfume)
particles move in constant, straight line motion
no forces between particles
Average Kinetic Energy is proportional to absolute temperature
Based on an Ideal Gas
an Ideal Gas has the following properties:
volume of each particle is essentially zero (perfume)
particles move in constant, straight line motion
no forces between particles
Average Kinetic Energy is proportional to absolute temperature
Kinetic Molecular TheoryKinetic Molecular TheoryKinetic Molecular TheoryKinetic Molecular Theory
Kinetic energy equation is KE = (1/2)mv2Kinetic energy equation is KE = (1/2)mv2
AtmosphereAtmosphereAtmosphereAtmosphere
Composed of 78% nitrogen, 21% oxygen, and <1% other gases
This atmospheric composition allows life to exist on Earth
Composed of 78% nitrogen, 21% oxygen, and <1% other gases
This atmospheric composition allows life to exist on Earth
Greenhouse gasesGreenhouse gasesGreenhouse gasesGreenhouse gases
Mainly caused by CO2 trapping heat from escaping our atmosphere
CO2 is produced from burning fossil fuels, clear cutting forests, as well as many other daily activities.
Mainly caused by CO2 trapping heat from escaping our atmosphere
CO2 is produced from burning fossil fuels, clear cutting forests, as well as many other daily activities.
OzoneOzoneOzoneOzone
Ozone (O3) blocks UV radiation from the sun
If all the ozone molecules in Earths atmosphere were stacked on earths surface, it would only form a 3 mm thick layer!
CFC’s break down O3 and is causing the ozone layer to disappear
Ozone (O3) blocks UV radiation from the sun
If all the ozone molecules in Earths atmosphere were stacked on earths surface, it would only form a 3 mm thick layer!
CFC’s break down O3 and is causing the ozone layer to disappear
Standard Temperature and Standard Temperature and Pressure(STP)Pressure(STP)
Standard Temperature and Standard Temperature and Pressure(STP)Pressure(STP)
STP is used to compare sets of data between gases
STP = 0 C and 1.00 atm
STP = 273.16 K and 760.00 mm Hg
STP = 273.16 K and 101.325 kPa
STP = 273.16 K and 760.0 torr
Molar Volume = 22.4 L
STP is used to compare sets of data between gases
STP = 0 C and 1.00 atm
STP = 273.16 K and 760.00 mm Hg
STP = 273.16 K and 101.325 kPa
STP = 273.16 K and 760.0 torr
Molar Volume = 22.4 L
Avogadro’s PrincipleAvogadro’s PrincipleAvogadro’s PrincipleAvogadro’s Principle
equal volumes of different gases under the same conditions have equal numbers of molecules
So 1.2 L of Carbon has the same number of molecules as 1.2 L of Chlorine
as mols increase so does volume
Remember amount is measured in mols and not grams!!!
equal volumes of different gases under the same conditions have equal numbers of molecules
So 1.2 L of Carbon has the same number of molecules as 1.2 L of Chlorine
as mols increase so does volume
Remember amount is measured in mols and not grams!!!
Gas ActivityGas ActivityGas ActivityGas Activity
I want each group to explain the following according to their situation
What happens to molecules
What happens to pressure
What happens to volume
What happens to temperature
I want each group to explain the following according to their situation
What happens to molecules
What happens to pressure
What happens to volume
What happens to temperature
Gas ActivityGas ActivityGas ActivityGas ActivityDraw on the SmartBoard and prepare a short presentation on what happens in your particular situation. Please include all relative information and draw a picture of what is happening
Table 1 - what happens to your balloon as you climb Mt. Everest
Table 2 - what happens to your balloon as you dive to the bottom of a deep sea trench
Table 3 - what happens to your balloon as you travel from the North Pole to the South Pole
Table 4 - what happens to your balloon as you travel around the equator (relatively same elevation, temperature, etc)
Table 5 - what happens to your balloon as you gradually fill it with helium until it explodes
Table 6 - what happens to your balloon as you take your balloon from an area of extremely high pressure and gradually into a vacuum (no pressure)
Draw on the SmartBoard and prepare a short presentation on what happens in your particular situation. Please include all relative information and draw a picture of what is happening
Table 1 - what happens to your balloon as you climb Mt. Everest
Table 2 - what happens to your balloon as you dive to the bottom of a deep sea trench
Table 3 - what happens to your balloon as you travel from the North Pole to the South Pole
Table 4 - what happens to your balloon as you travel around the equator (relatively same elevation, temperature, etc)
Table 5 - what happens to your balloon as you gradually fill it with helium until it explodes
Table 6 - what happens to your balloon as you take your balloon from an area of extremely high pressure and gradually into a vacuum (no pressure)
Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s Law
• The pressure - volume relationship• The pressure - volume relationship
Cartesian DiverCartesian DiverCartesian DiverCartesian Diver
How does this work?How does this work?
Pop GunPop GunPop GunPop Gun
How does this work?How does this work?
Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s Law
Imagine a syringe - as you compress the plunger the trapped gas particles collide with the walls more often which results in higher pressure
If you push with a fixed pressure, the volume will decrease until the pressure inside equals the pressure outside
Temperature must be kept constant
Imagine a syringe - as you compress the plunger the trapped gas particles collide with the walls more often which results in higher pressure
If you push with a fixed pressure, the volume will decrease until the pressure inside equals the pressure outside
Temperature must be kept constant
Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s Law
Boyle’s Law states that pressure and volume are inversely proportional
PV = k (constant)
More commonly seen as P = 1/V or V = 1/P
P1V1 = P2V2 when T is constant
Boyle’s Law states that pressure and volume are inversely proportional
PV = k (constant)
More commonly seen as P = 1/V or V = 1/P
P1V1 = P2V2 when T is constant
Sample ProblemSample ProblemSample ProblemSample ProblemThe gas in a balloon has a volume of 4L at 100kPa. The balloon is released into the atmosphere, and the gas in it expands to a volume of 8L. What is the pressure on the balloon at the new volume?
What do we know?
V1 = 4L
V2 = 8L
P1 = 100kPa
P2 = ? (unknown)
The gas in a balloon has a volume of 4L at 100kPa. The balloon is released into the atmosphere, and the gas in it expands to a volume of 8L. What is the pressure on the balloon at the new volume?
What do we know?
V1 = 4L
V2 = 8L
P1 = 100kPa
P2 = ? (unknown)
Sample Problem Cont’Sample Problem Cont’Sample Problem Cont’Sample Problem Cont’
Use equation P1V1 = P2V2
Rearrange to solve for unknown P2=P1V1/V2
Plug in values
P2 = (100)(4)/8
P2 = 50kPa
Use equation P1V1 = P2V2
Rearrange to solve for unknown P2=P1V1/V2
Plug in values
P2 = (100)(4)/8
P2 = 50kPa
Sample ProblemSample ProblemSample ProblemSample Problem
If the pressure of a 2.5 m3 sample of a gas is 1.5 atm, what volume will the gas occupy if the pressure is changed to 7.5 atm
What do we know?
P1 = 1.5
V1 = 2.5
P2 = 7.5
V2 = ?
If the pressure of a 2.5 m3 sample of a gas is 1.5 atm, what volume will the gas occupy if the pressure is changed to 7.5 atm
What do we know?
P1 = 1.5
V1 = 2.5
P2 = 7.5
V2 = ?
Sample Problem Cont’Sample Problem Cont’Sample Problem Cont’Sample Problem Cont’
Use P1V1 = P2V2
Rearrange equation to solve for what we want V2 = P1V1/P2
Plug in variables
V2 = (1.5 X 2.5)/7.5
V2 = .5
Use P1V1 = P2V2
Rearrange equation to solve for what we want V2 = P1V1/P2
Plug in variables
V2 = (1.5 X 2.5)/7.5
V2 = .5
Charles's LawCharles's LawCharles's LawCharles's Law
• The temperature vs. volume relationship
• The temperature vs. volume relationship
Egg DemoEgg DemoEgg DemoEgg Demo
How does this work?How does this work?
VideoVideoVideoVideo
Better demo of balloon exampleBetter demo of balloon example
QuickTime™ and a decompressor
are needed to see this picture.
Crunching Can Crunching Can Crunching Can Crunching Can
How does this work?How does this work?
Charles’s LawCharles’s LawCharles’s LawCharles’s Law
Particles move faster when they are heated
Particles push with more force when they are heated
Particles move faster when they are heated
Particles push with more force when they are heated
Charles’s LawCharles’s LawCharles’s LawCharles’s Law
Temperature and Volume are directly proportional, so as temp in/decreases, so does volume.
Temperature and Volume are directly proportional, so as temp in/decreases, so does volume.
Absolute Temperature Absolute Temperature ScaleScaleAbsolute Temperature Absolute Temperature ScaleScale
Measured in Kelvins
Kelvins are just like ˚C or ˚F
0˚C = 273 K
Measured in Kelvins
Kelvins are just like ˚C or ˚F
0˚C = 273 K
Kelvin Temperature ScaleKelvin Temperature ScaleKelvin Temperature ScaleKelvin Temperature Scale
Converting ˚C to K = take ˚C temp and add 273
Converting K to ˚C = take K temp and subtract 273
0˚ K is theoretically lowest possible temperature
Converting ˚C to K = take ˚C temp and add 273
Converting K to ˚C = take K temp and subtract 273
0˚ K is theoretically lowest possible temperature
Charles’s Law Charles’s Law Charles’s Law Charles’s Law
Charles’s Law is Volume/Temperature or V/T
It is also used as V1/T1=V2/T2 when Pressure is constant
It is also used as V1T2=T1V2
Charles’s Law is Volume/Temperature or V/T
It is also used as V1/T1=V2/T2 when Pressure is constant
It is also used as V1T2=T1V2
Sample ProblemSample ProblemSample ProblemSample Problem
A sample of gas occupies 24 m3 at 100K. What volume would the gas occupy at 400K?
What do we know?
V1 = 24
T1 = 100
T2 = 400
V2 = ?
A sample of gas occupies 24 m3 at 100K. What volume would the gas occupy at 400K?
What do we know?
V1 = 24
T1 = 100
T2 = 400
V2 = ?
Sample Problem Cont’Sample Problem Cont’Sample Problem Cont’Sample Problem Cont’
Charles Law is V1/T1 = V2/T2
Rearrange it to V2 = (V1)(T2)/T1
Plug in what you know
V2 = (24)(400)/100
V2 = 96 m3
Charles Law is V1/T1 = V2/T2
Rearrange it to V2 = (V1)(T2)/T1
Plug in what you know
V2 = (24)(400)/100
V2 = 96 m3
Sample ProblemSample ProblemSample ProblemSample Problem
Gas in a balloon occupies 2.5 L at 54˚C at what temperature will the balloon expand to 7.5 L?
What do we know?
V1 = 2.5
T1 = 54˚C
V2 = 7.5
T2 = ?
Gas in a balloon occupies 2.5 L at 54˚C at what temperature will the balloon expand to 7.5 L?
What do we know?
V1 = 2.5
T1 = 54˚C
V2 = 7.5
T2 = ?
Sample Problem Cont’Sample Problem Cont’Sample Problem Cont’Sample Problem Cont’
Equation = V1/T1 = V2/T2
Rearrange = T2 = (T1)(V2)/V1
Convert 54˚C to K (54 + 273 = 327)
Plug in what you know
T2 = (327)(7.5)/2.5
T2 = 980 K (980-273) = 707 ˚C
Equation = V1/T1 = V2/T2
Rearrange = T2 = (T1)(V2)/V1
Convert 54˚C to K (54 + 273 = 327)
Plug in what you know
T2 = (327)(7.5)/2.5
T2 = 980 K (980-273) = 707 ˚C
Converting from ˚C to KConverting from ˚C to KConverting from ˚C to KConverting from ˚C to K
Remember K = ˚C + 273
37˚C
2˚C
150˚C
84˚C
534˚C
Remember K = ˚C + 273
37˚C
2˚C
150˚C
84˚C
534˚C
Converting from K to ˚CConverting from K to ˚CConverting from K to ˚CConverting from K to ˚C
Remember ˚C = K - 273
465 K
273 K
487 K
182 K
65 K
Remember ˚C = K - 273
465 K
273 K
487 K
182 K
65 K
No Name LawNo Name LawThe Pressure - Temperature RelationshipThe Pressure - Temperature Relationship
No Name LawNo Name LawThe Pressure - Temperature RelationshipThe Pressure - Temperature Relationship
Also known as the Gay-Lussac Law
Gay-Lussac used unpublished info from Charles Law and made the following law:
P1/T1 = P2/T2
This Law is also used as P1T2 = P2T1
We will use this law in the lab
Remember temperature must be in Kelvins
Also known as the Gay-Lussac Law
Gay-Lussac used unpublished info from Charles Law and made the following law:
P1/T1 = P2/T2
This Law is also used as P1T2 = P2T1
We will use this law in the lab
Remember temperature must be in Kelvins
Combined Gas LawCombined Gas LawCombined Gas LawCombined Gas Law
• The pressure - volume - temperature relationship
• The pressure - volume - temperature relationship
Combined Gas LawCombined Gas LawCombined Gas LawCombined Gas Law
Boyles Law + Charles Law + Gay-Lussacs Law = Combined Gas LawBoyles Law + Charles Law + Gay-Lussacs Law = Combined Gas Law
Boyles LawBoyles LawBoyles LawBoyles Law
Boyle’s Law states that pressure and volume are _________ proportional
Boyles Law equation = __________
(P1)(V1) = (P2)(V2) or P = 1/V and V = 1/P
Temperature is constant
Boyle’s Law states that pressure and volume are _________ proportional
Boyles Law equation = __________
(P1)(V1) = (P2)(V2) or P = 1/V and V = 1/P
Temperature is constant
Charles LawCharles LawCharles LawCharles Law
Charles’s Law states that volume and temperature are ________ proportional
Charles Law equation = ______
(V1)(T2) = (T1)(V2) or V1/T1 = V2/T2
Pressure is constant
Charles’s Law states that volume and temperature are ________ proportional
Charles Law equation = ______
(V1)(T2) = (T1)(V2) or V1/T1 = V2/T2
Pressure is constant
Combined Gas LawCombined Gas LawCombined Gas LawCombined Gas Law
This law combines all of the laws into one equation
P1V1/T1 = P2V2/T2
Or (P1)(V1)(T2) = (P2)(V2)(T1)
The number of molecules is constant
Temperature must be in Kelvins
This law combines all of the laws into one equation
P1V1/T1 = P2V2/T2
Or (P1)(V1)(T2) = (P2)(V2)(T1)
The number of molecules is constant
Temperature must be in Kelvins
Combined Gas Law GameCombined Gas Law GameCombined Gas Law GameCombined Gas Law Game
Go to the following address and read the directions, then play the game
The purpose of the game is to keep the molecules constant
You need to vary the temperature, to increase or decrease pressure and volume
http://www.chem11games.net
Go to the following address and read the directions, then play the game
The purpose of the game is to keep the molecules constant
You need to vary the temperature, to increase or decrease pressure and volume
http://www.chem11games.net
Sample ProblemSample ProblemSample ProblemSample Problem
A helium balloon with a volume of 410 mL is cooled from 27˚C to -27˚C. The pressure on the gas is reduced from 110 kPa to 25 kPa. What is the volume of the gas at the lower temperature and pressure?
A helium balloon with a volume of 410 mL is cooled from 27˚C to -27˚C. The pressure on the gas is reduced from 110 kPa to 25 kPa. What is the volume of the gas at the lower temperature and pressure?
Sample Problem Cont’...Sample Problem Cont’...Sample Problem Cont’...Sample Problem Cont’...
What do we know?
P1 = 110 kPa
V1 = 410 mL
T1 = 27˚C + 273 = 300K
P2 = 25 kPa
V2 = ?
T2 = -27˚C = 246 K
What do we know?
P1 = 110 kPa
V1 = 410 mL
T1 = 27˚C + 273 = 300K
P2 = 25 kPa
V2 = ?
T2 = -27˚C = 246 K
Sample Problem Cont’...Sample Problem Cont’...Sample Problem Cont’...Sample Problem Cont’...
Plug in what you know
(P1)(V1) / T1 = (P2)(V2) / T2
(110)(410) / 300 = (25)(V2) / 246
Solve:
150.3 = (25)(V2) / 246
36982 = (25)(V2)
1480 mL = V2
Plug in what you know
(P1)(V1) / T1 = (P2)(V2) / T2
(110)(410) / 300 = (25)(V2) / 246
Solve:
150.3 = (25)(V2) / 246
36982 = (25)(V2)
1480 mL = V2
Sample ProblemSample ProblemSample ProblemSample Problem
A 350 cm3 sample of helium gas is collected at 22.0 ˚C and 99.3 kPa. What volume would this gas occupy at STP?
A 350 cm3 sample of helium gas is collected at 22.0 ˚C and 99.3 kPa. What volume would this gas occupy at STP?
What do we know?
V1 = 350 cm3
P1 = 99.3 kPa
T1 = 22 ˚C
V2 = ?
P2 = 101.3 kPa
T2 = 273 K
What do we know?
V1 = 350 cm3
P1 = 99.3 kPa
T1 = 22 ˚C
V2 = ?
P2 = 101.3 kPa
T2 = 273 K
Convert temperature into Kelvins 22˚C = _____
Plug in values
(P1)(V1)(T2) = (P2)(V2)(T1)
(99.3)(350)(273) = (101.3)(V2)(295)
Solve:
9,488,115 = 29,883.5 (V2)
318 cm3 = V2
Convert temperature into Kelvins 22˚C = _____
Plug in values
(P1)(V1)(T2) = (P2)(V2)(T1)
(99.3)(350)(273) = (101.3)(V2)(295)
Solve:
9,488,115 = 29,883.5 (V2)
318 cm3 = V2
Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law
• All Four Variables• All Four Variables
Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law
•PV = nRT
•(Pressure)(Volume) = (Moles)(R-constant)(Temperature)
•PV = nRT
•(Pressure)(Volume) = (Moles)(R-constant)(Temperature)
R - constantR - constantR - constantR - constant
•R is calculated differently depending upon the units of pressure
• If you are using kPa then R = 8.314 (L)(kPa)/(mol)(K)
• If you are using atm then R = .0821 (L)(kPa)/(mol)(K)
•R is calculated differently depending upon the units of pressure
• If you are using kPa then R = 8.314 (L)(kPa)/(mol)(K)
• If you are using atm then R = .0821 (L)(kPa)/(mol)(K)
Solving ProblemsSolving ProblemsSolving ProblemsSolving Problems
•Temperature must be in kelvins
•Volume must be in liters
•Pressure must be in kPa or atm
•Temperature must be in kelvins
•Volume must be in liters
•Pressure must be in kPa or atm
Ideal Gas Sample ProblemIdeal Gas Sample ProblemIdeal Gas Sample ProblemIdeal Gas Sample Problem
•A sample of carbon dioxide with a mass of .250g was placed in a 350mL container at 400K. What is the pressure exerted by the gas?
•A sample of carbon dioxide with a mass of .250g was placed in a 350mL container at 400K. What is the pressure exerted by the gas?
Ideal Gas Sample ProblemIdeal Gas Sample ProblemIdeal Gas Sample ProblemIdeal Gas Sample Problem
•What do we know?
•P = ?
•V = 350mL
•T = 400K
•n = calculated from .250g and 44.01 molar mass
•R = 8.314 LkPa/molK
•What do we know?
•P = ?
•V = 350mL
•T = 400K
•n = calculated from .250g and 44.01 molar mass
•R = 8.314 LkPa/molK
Ideal Gas Sample ProblemIdeal Gas Sample ProblemIdeal Gas Sample ProblemIdeal Gas Sample Problem
•Calculate n = .250g X 1 mol / 44.01g = .00568
•Plug in values
•PV = nRT
•(P)(.35L) = (.00568)(8.31)(400)
•P = 54 kPa
•Calculate n = .250g X 1 mol / 44.01g = .00568
•Plug in values
•PV = nRT
•(P)(.35L) = (.00568)(8.31)(400)
•P = 54 kPa
Gas StoichiometryGas StoichiometryGas StoichiometryGas Stoichiometry
•Must use mole ratio
•Remember mole ratio means “what you need over what you have”
•Need/Have
•Must use mole ratio
•Remember mole ratio means “what you need over what you have”
•Need/Have
Calculating RatioCalculating RatioCalculating RatioCalculating Ratio
•What is the ratio if the problem says, how many moles of hydrogen would react with 4 mol of nitrogen according to the following equation?
•3H2 + N2 ----- 2NH3
•What do we need?
•What do we have?
•So the ratio is 3/1
•What is the ratio if the problem says, how many moles of hydrogen would react with 4 mol of nitrogen according to the following equation?
•3H2 + N2 ----- 2NH3
•What do we need?
•What do we have?
•So the ratio is 3/1
Example 2Example 2Example 2Example 2
•How many moles of oxygen would react with 2 mol of H2 according to the following equation?
•2H2 + O2 ---- 2H2O
•What do we need?
•What do we have?
•So the ratio is 1/2
•How many moles of oxygen would react with 2 mol of H2 according to the following equation?
•2H2 + O2 ---- 2H2O
•What do we need?
•What do we have?
•So the ratio is 1/2
Gas Stoich Sample problemGas Stoich Sample problemGas Stoich Sample problemGas Stoich Sample problem
•How many liters of hydrogen gas will be produced at 280K and 96kPa if 40 g of sodium react with excess hydrochloric acid?
•Step 1 - write out reaction 2Na + 2HCl ----- 2NaCl + H2
•What is ratio?
•How many liters of hydrogen gas will be produced at 280K and 96kPa if 40 g of sodium react with excess hydrochloric acid?
•Step 1 - write out reaction 2Na + 2HCl ----- 2NaCl + H2
•What is ratio?
Gas Stoich Sample ProblemGas Stoich Sample ProblemGas Stoich Sample ProblemGas Stoich Sample Problem
•Step 2 - List Knowns
•V = ?
•P = 96 kPa
•n = calculated from grams, molar mass, and ratio
•Molar Mass = Na = 23 g/mol
•R = 8.314 LkPa/molK
•T = 280K
•Step 2 - List Knowns
•V = ?
•P = 96 kPa
•n = calculated from grams, molar mass, and ratio
•Molar Mass = Na = 23 g/mol
•R = 8.314 LkPa/molK
•T = 280K
Gas Stoich SampleGas Stoich SampleGas Stoich SampleGas Stoich Sample
•Step 3 - Set up
•convert grams to moles - 40g X 1mol/23g = 1.74 mol Na
•multiply by ratio - 1.74 mol Na X 1/2 = .870 mol H2
•Plug into equation - PV = nRT
•Solve:
•(96)(V) = (.870)(8.314)(280)
•(96)(V) = 2025.3
•V = 21.1L of H2
•Step 3 - Set up
•convert grams to moles - 40g X 1mol/23g = 1.74 mol Na
•multiply by ratio - 1.74 mol Na X 1/2 = .870 mol H2
•Plug into equation - PV = nRT
•Solve:
•(96)(V) = (.870)(8.314)(280)
•(96)(V) = 2025.3
•V = 21.1L of H2