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

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Presentation on theme: "Ch. 14 - Gases II. Ideal Gas Law Ideal Gas Law and Gas Stoichiometry."— Presentation transcript:

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

2 Part 1 Ideal Gas Law

3 1 mol of a gas=22.4 L at STP Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm A. Avogadros Principle

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

5 PV T VnVn PV nT B. Ideal Gas Law = k UNIVERSAL GAS CONSTANT R= L atm/mol K R=8.315 dm 3 kPa/mol K = R Merge the Combined Gas Law with Avogadros Principle:

6 B. Ideal Gas Law UNIVERSAL GAS CONSTANT R= L atm/mol K R=8.315 dm 3 kPa/mol K PV=nRT

7 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 b Calculate the pressure in atmospheres of mol of He at 16°C & occupying 3.25 L.

8 GIVEN: V = ?V = ? n = 85 g T = 25°C = 298 K P = kPa R = dm 3 kPa/mol K C. Ideal Gas Law Problems b Find the volume of 85 g of O 2 at 25°C and kPa. = 2.7 mol WORK: 85 g 1 mol O 2 = 2.7 mol gO 2 PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3 kPa/mol K K V = 64 dm 3

9 D. Applications of Ideal Gas Law b Can be used to calculate the molar mass of a gas from the density b Substitute this into ideal gas law b And m/V = d in g/L, so

10 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 b 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.

11 GIVEN: d = ? g/L CO 2 T = 25°C = 298 K P = 750. torr R = L atm/mol K MM = g/mol MM = dRT/P d = MM P/RT d=(44.01 g/mol )(.987 atm ) ( L atm/mol K )(298 K ) d = 1.78 g/L CO 2 D. Applications of Ideal Gas Law b Calculate the density of carbon dioxide gas at 25 °C and 750. torr. WORK: 750 torr 1 atm =.987 atm 760 torr =.987 atm

12 Part 2 Gas Stoichiometry

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

14 1 mol of a gas=22.4 L at STP A. Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm

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

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

17 C. Gas Stoichiometry - Volume b What volume of oxygen is needed for the complete combustion of 4.00 L of propane (C 3 H 8 )? C 3 H 8 + O 2 CO 2 + H 2 O L ?L 4.00 L C 3 H 8 5 L O 2 1 L C 3 H 8 = 20.0 L O 2

18 b How many grams of KClO 3 are reqd to produce 9.00 L of O 2 at STP? 9.00 L O 2 1 mol O L O 2 = 32.8 g KClO 3 2 mol KClO 3 3 mol O g KClO 3 1 mol KClO 3 ? g9.00 L C. Gas Stoichiometry Problem – STP 2KClO 3 2KCl + 3O 2

19 1 mol CaCO g CaCO 3 D. Gas Stoichiometry Problem – Non-STP b What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? 5.25 g CaCO 3 = mol CO 2 CaCO 3 CaO + CO 2 1 mol CO 2 1 mol CaCO g? L non-STP Looking for liters: Start with stoich and calculate moles of CO 2. Plug this into the Ideal Gas Law for n to find liters NEXT P = 103 kPa V = ? n = ? R = dm 3 kPa/molK T = 298K

20 WORK: PV = nRT (103 kPa)V =(0.0525mol)(8.315 dm 3 kPa/mol K ) (298K) V = 1.26 dm 3 CO 2 b What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = mol T = 25°C = 298 K R = dm 3 kPa/mol K D. Gas Stoichiometry Problem – Non-STP

21 WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315 dm 3 kPa/mol K ) (294K) n = mol O 2 B. Gas Stoichiometry Problem b How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? GIVEN: P = 97.3 kPa V = 15.0 L n = ?n = ? T = 21°C = 294 K R = dm 3 kPa/mol K 4 Al + 3 O 2 2 Al 2 O L non-STP ? g Given liters: Start with Ideal Gas Law and calculate moles of O 2. NEXT

22 2 mol Al 2 O 3 3 mol O 2 B. Gas Stoichiometry Problem b How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? mol O 2 = 40.6 g Al 2 O 3 4 Al + 3 O 2 2 Al 2 O g Al 2 O 3 1 mol Al 2 O L non-STP ? g Use stoich to convert moles of O 2 to grams Al 2 O 3


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