Introduction to 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