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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
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Part 1Part 1Part 1Part 1
Ideal Gas Law
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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
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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
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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:
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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
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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.
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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
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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
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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.
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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
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Part 2Part 2Part 2Part 2
Gas Stoichiometry
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* 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.
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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
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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
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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
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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
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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
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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
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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
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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
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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