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

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

Part 1 Ideal Gas Law

Standard Temperature & Pressure
A. Avogadro’s Principle Molar Volume at STP 1 mol of a gas=22.4 L at STP Standard Temperature & Pressure 0°C and 1 atm

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

UNIVERSAL GAS CONSTANT
B. Ideal Gas Law Merge the Combined Gas Law with Avogadro’s Principle: V n PV nT PV T = k = R UNIVERSAL GAS CONSTANT R= Latm/molK R=8.315 dm3kPa/molK

UNIVERSAL GAS CONSTANT
B. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R= Latm/molK R=8.315 dm3kPa/molK

C. Ideal Gas Law Problems
Calculate the pressure in atmospheres of mol of He at 16°C & occupying L. GIVEN: P = ? atm n = mol T = 16°C = 289 K V = 3.25 L R = Latm/molK WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K P = 3.01 atm

C. Ideal Gas Law Problems
Find the volume of 85 g of O2 at 25°C and kPa. GIVEN: V = ? n = 85 g T = 25°C = 298 K P = kPa R = dm3kPa/molK WORK: 85 g 1 mol O2 = 2.7 mol 32.00 g O2 = 2.7 mol PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K V = 64 dm3

D. Applications of Ideal Gas Law
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

D. Applications of Ideal Gas Law
The density of a gas was measured at atm and 27°C and found to be 1.95 g/L. Calculate the molar mass of the gas. GIVEN: P = 1.50 atm T = 27°C = 300. K d = 1.95 g/L R = Latm/molK MM = ? WORK: MM = dRT/P MM=(1.95)( )(300.)/1.50 g/L Latm/molK K atm MM = 32.0 g/mol

D. Applications of Ideal Gas Law
Calculate the density of carbon dioxide gas at 25°C and 750. torr. GIVEN: d = ? g/L CO2 T = 25°C = 298 K P = 750. torr R = Latm/molK MM = g/mol WORK: 750 torr 1 atm = .987 atm torr MM = dRT/P →d = MM P/RT d=(44.01 g/mol)(.987 atm) ( Latm/molK )(298K) d = 1.78 g/L CO2 = .987 atm

Part 2 Gas Stoichiometry

* Stoichiometry Steps 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 Mole ratio - moles  moles Core step in all stoichiometry problems!! 4. Check answer.

Standard Temperature & Pressure
A. Molar Volume at STP 1 mol of a gas=22.4 L at STP Standard Temperature & Pressure 0°C and 1 atm

A. Molar Volume at STP LITERS OF GAS AT STP Molar Volume MASS IN GRAMS
(22.4 L/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Mass (g/mol) 6.02  1023 particles/mol

B. Gas Stoichiometry Liters of one Gas  Liters of another Gas:
Avogadro’s Principle Coefficients give mole ratios and volume ratios Moles (or grams) of A  Liters of B: STP – use 22.4 L/mol Non-STP – use ideal gas law & stoich Non-STP Given liters of gas? start with ideal gas law Looking for liters of gas? start with stoichiometry conv

C. Gas Stoichiometry - Volume
What volume of oxygen is needed for the complete combustion of 4.00 L of propane (C3H8)? C3H O2  CO2+ H2O 4.00 L ?L 5 L O2 1 L C3H8 4.00 L C3H8 = 20.0 L O2

C. Gas Stoichiometry Problem – STP
How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP? 2KClO3  2KCl + 3O2 ? g 9.00 L 9.00 L O2 1 mol O2 22.4 L 2 mol KClO3 3 mol O2 122.55 g KClO3 1 mol KClO3 = 32.8 g KClO3

D. Gas Stoichiometry Problem – Non-STP
What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC? CaCO3  CaO CO2 5.25 g ? L non-STP Looking for liters: Start with stoich and calculate moles of CO2. P = 103 kPa V = ? n = ? R = dm3kPa/molK T = 298K 5.25 g CaCO3 1 mol CaCO3 100.09g 1 mol CO2 CaCO3 NEXT  = mol CO2 Plug this into the Ideal Gas Law for n to find liters

D. Gas Stoichiometry Problem – Non-STP
What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = mol T = 25°C = 298 K R = dm3kPa/molK WORK: PV = nRT (103 kPa)V =(0.0525mol)(8.315dm3kPa/molK) (298K) V = 1.26 dm3 CO2

B. Gas Stoichiometry Problem
How many grams of Al2O3 are formed from L of O2 at 97.3 kPa & 21°C? 4 Al O2  2 Al2O3 15.0 L non-STP ? g GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = dm3kPa/molK WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315dm3kPa/molK) (294K) n = mol O2 Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT 

B. Gas Stoichiometry Problem
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al O2  2 Al2O3 15.0L non-STP ? g Use stoich to convert moles of O2 to grams Al2O3 0.597 mol O2 2 mol Al2O3 3 mol O2 g Al2O3 1 mol Al2O3 = 40.6 g Al2O3

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