Ch. 14 - Gases II. Ideal Gas Law Ideal Gas Law and Gas Stoichiometry.

Ch. 14 - Gases

II. Ideal Gas II. Ideal Gas LawLawII. Ideal Gas II. Ideal Gas LawLaw

Ideal Gas Law and Gas Ideal Gas Law and Gas StoichiometryStoichiometry

Part 1Part 1Part 1Part 1

Ideal Gas Law

1 mol of a gas=22.4 Lat STP

Molar Volume at Molar Volume at STPSTPMolar Volume at Molar Volume at STPSTP

Standard Temperature & Pressure0°C and 1 atm

A. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s Principle

kn

VV

n

A. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s Principle

Equal volumes of gases contain equal numbers of moles• at constant temp & pressure• true for any gas• n = number of moles

PV

TVn

PVnT

B. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas Law

= kUNIVERSAL GAS

CONSTANTR=0.08206

Latm/molKR=8.315

dm3kPa/molK

= R

Merge the Combined Gas Law with Avogadro’s Principle:

B. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas Law

UNIVERSAL GAS CONSTANTR=0.08206

Latm/molKR=8.315

dm3kPa/molK

PV=nRT

GIVEN:

P = ? atm

n = 0.412 mol

T = 16°C = 289 K

V = 3.25 LR = 0.08206Latm/molK

WORK:

PV = nRT

P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K

P = 3.01 atm

C. Ideal Gas Law ProblemsC. Ideal Gas Law ProblemsC. Ideal Gas Law ProblemsC. Ideal Gas Law Problems Calculate the pressure in atmospheres of

0.412 mol of He at 16°C & occupying 3.25 L.

GIVEN:

V = ?

n = 85 g

T = 25°C = 298 K

P = 104.5 kPaR = 8.315 dm3kPa/molK

C. Ideal Gas Law ProblemsC. Ideal Gas Law ProblemsC. Ideal Gas Law ProblemsC. Ideal Gas Law Problems

Find the volume of 85 g of O2 at 25°C and 104.5 kPa.

= 2.7 mol

WORK:

85 g 1 mol O2 = 2.7 mol

32.00 g O2

PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K

V = 64 dm3

D. Applications of Ideal Gas D. Applications of Ideal Gas LawLawD. Applications of Ideal Gas D. Applications of Ideal Gas LawLaw

Can be used to calculate the molar mass of a gas from the density

Substitute this into ideal gas law

And m/V = d in g/L, so

MMmassmolar

mass

massmolar

gas a of grams gas a of moles

mn

)(

)(

VMM

RTm

V

nRTP

P

dRTMM

MM

dRTP or

GIVEN:

P = 1.50 atm

T = 27°C = 300. K

d = 1.95 g/LR = 0.08206 Latm/molK

MM = ?

WORK:

MM = dRT/P

MM=(1.95)(0.08206)(300.)/1.50 g/L Latm/molK K atm

MM = 32.0 g/mol

D. Applications of Ideal Gas D. Applications of Ideal Gas LawLawD. Applications of Ideal Gas D. Applications of Ideal Gas LawLaw

The density of a gas was measured at 1.50 atm and 27°C and found to be 1.95 g/L. Calculate the molar mass of the gas.

GIVEN:

d = ? g/L CO2

T = 25°C = 298 KP = 750. torr

R = 0.08206 Latm/molK

MM = 44.01 g/mol

MM = dRT/P →d = MM P/RT

d=(44.01 g/mol)(.987 atm)

(0.08206 Latm/molK )(298K)

d = 1.78 g/L CO2

D. Applications of Ideal Gas D. Applications of Ideal Gas LawLawD. Applications of Ideal Gas D. Applications of Ideal Gas LawLaw

Calculate the density of carbon dioxide gas at 25°C and 750. torr.

WORK:

750 torr 1 atm = .987 atm 760 torr

= .987 atm

Part 2Part 2Part 2Part 2

Gas Stoichiometry

* Stoichiometry Steps * Stoichiometry Steps Review *Review ** Stoichiometry Steps * Stoichiometry Steps Review *Review *

1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.

• Mole ratio - moles moles• Molar mass - moles grams• Molarity - moles liters soln• Molar volume - moles liters gas

Core step in all stoichiometry problems!!

• Mole ratio - moles moles

4. Check answer.

1 mol of a gas=22.4 Lat STP

A. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STP

Standard Temperature & Pressure0°C and 1 atm

A. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STP

Molar Mass(g/mol)

6.02 1023

particles/mol

MASSIN

GRAMSMOLES

NUMBEROF

PARTICLES

Molar Volume (22.4 L/mol)

LITERSOF GASAT STP

B. Gas StoichiometryB. Gas StoichiometryB. Gas StoichiometryB. Gas Stoichiometry

Liters of one Gas Liters of one Gas Liters of another Gas: Liters of another Gas:• Avogadro’s Principle • Coefficients give mole ratios and volume

ratios Moles (or grams) of A Moles (or grams) of A Liters of B: Liters of B:

• STP – use 22.4 L/mol • Non-STP – use ideal gas law & stoich

Non-Non-STPSTP• Given liters of gas?

start with ideal gas law• Looking for liters of gas?

start with stoichiometry conv

C. Gas Stoichiometry - C. Gas Stoichiometry - VolumeVolumeC. Gas Stoichiometry - C. Gas Stoichiometry - VolumeVolume

What volume of oxygen is needed for the complete combustion of 4.00 L of propane (C3H8)?

C3H8 + O2 CO2+ H2O5 3 44.00 L ?L

4.00 L C3H85 L O2

1 L C3H8

= 20.0 L O2

How many grams of KClO3 are req’d to

produce 9.00 L of O2 at STP?

9.00 LO2

1 molO2

22.4 L O2

= 32.8 g KClO3

2 molKClO3

3 molO2

122.55g KClO3

1 molKClO3

? g 9.00 L

C. Gas Stoichiometry Problem – C. Gas Stoichiometry Problem – STPSTPC. Gas Stoichiometry Problem – C. Gas Stoichiometry Problem – STPSTP

2KClO3 2KCl + 3O2

1 molCaCO3

100.09g CaCO3

D. Gas Stoichiometry Problem – Non-D. Gas Stoichiometry Problem – Non-STPSTPD. Gas Stoichiometry Problem – Non-D. Gas Stoichiometry Problem – Non-STPSTP

What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?

5.25 gCaCO3 = 0.0525 mol CO2

CaCO3 CaO + CO2

1 molCO2

1 molCaCO3

5.25 g ? Lnon-STPLooking for liters: Start with stoich

and calculate moles of CO2.

Plug this into the Ideal Gas Law for n to find liters

NEXT P = 103 kPa

V = ?

n = ?

R = 8.315 dm3kPa/molK

T = 298K

WORK:

PV = nRT

(103 kPa)V=(0.0525mol)(8.315dm3kPa/molK) (298K)

V = 1.26 dm3 CO2

What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?

GIVEN:

P = 103 kPaV = ?

n = 0.0525 molT = 25°C = 298 KR = 8.315 dm3kPa/molK

D. Gas Stoichiometry Problem – Non-D. Gas Stoichiometry Problem – Non-STPSTPD. Gas Stoichiometry Problem – Non-D. Gas Stoichiometry Problem – Non-STPSTP

WORK:

PV = nRT

(97.3 kPa) (15.0 L)= n (8.315dm3kPa/molK) (294K)

n = 0.597 mol O2

B. Gas Stoichiometry B. Gas Stoichiometry ProblemProblemB. Gas Stoichiometry B. Gas Stoichiometry ProblemProblem

How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?

GIVEN:

P = 97.3 kPaV = 15.0 L

n = ?T = 21°C = 294 KR = 8.315 dm3kPa/molK

4 Al + 3 O2 2 Al2O3 15.0 L

non-STP ? gGiven liters: Start with

Ideal Gas Law and calculate moles of O2.

NEXT

2 mol Al2O3

3 mol O2

B. Gas Stoichiometry B. Gas Stoichiometry ProblemProblemB. Gas Stoichiometry B. Gas Stoichiometry ProblemProblem

How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?

0.597mol O2 = 40.6 g Al2O3

4 Al + 3 O2 2 Al2O3

101.96 g Al2O3

1 molAl2O3

15.0Lnon-STP

? gUse stoich to convert moles of O2 to grams Al2O3