Ch 14 Gases II Ideal Gas Law and

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

Part 1 Ideal Gas Law

A. Avogadro’s Principle Molar Volume at STP 1 mol of a gas=22. 4 L at STP Standard Temperature & 0°C and 1 atm Pressure

A. Avogadro’s Principle b V n Equal volumes of gases contain equal numbers of moles • at constant temp & pressure • true for any gas • n = number of moles

B. Ideal Gas Law Merge the Combined Gas Law with Avogadro’s Principle: PV V =R k n. T T n UNIVERSAL GAS CONSTANT R=0. 08206 L atm/mol K R=8. 315 dm 3 k. Pa/mol K

B. Ideal Gas Law PV=n. RT UNIVERSAL GAS CONSTANT R=0. 08206 L atm/mol K R=8. 315 dm 3 k. Pa/mol K

C. Ideal Gas Law Problems b Calculate the pressure in atmospheres of 0. 412 mol of He at 16°C & occupying 3. 25 L. GIVEN: WORK: P = ? atm PV = n. RT n = 0. 412 mol P(3. 25)=(0. 412)(0. 0821)(289) mol L atm/mol K T = 16°C = 289 K K L V = 3. 25 L P = 3. 01 atm R= 0. 08206 L atm/mol K

C. Ideal Gas Law Problems b Find the volume of 85 g of O 2 at 25°C and 104. 5 k. Pa. GIVEN: WORK: V=? 85 g 1 mol O 2 = 2. 7 mol n = 85 g = 2. 7 mol 32. 00 g O 2 T = 25°C = 298 K PV = n. RT P = 104. 5 k. Pa (104. 5)V=(2. 7) (8. 315) (298) k. Pa mol dm 3 k. Pa/mol K R = 8. 315 V = 64 dm 3 k. Pa/mol K

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

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. GIVEN: WORK: MM = d. RT/P P = 1. 50 atm T = 27°C = 300. K MM=(1. 95)(0. 08206)(300. )/1. 50 g/L L atm/mol K K d = 1. 95 g/L atm R = 0. 08206 MM = 32. 0 g/mol L atm/mol K MM = ?

D. Applications of Ideal Gas Law b. Calculate the density of carbon dioxide gas at 25°C and 750. torr. GIVEN: WORK: 750 torr 1 atm =. 987 atm d = ? g/L CO 2 760 torr T = 25°C = 298 K MM = d. RT/P →d = MM P/RT P = 750. torr =. 987 atm d=(44. 01 g/mol)(. 987 atm) R = 0. 08206 L atm/mol K MM = 44. 01 g/mol (0. 08206 L atm/mol K )(298 K) d = 1. 78 g/L CO 2

Part 2 Gas Stoichiometry

* Stoichiometry Steps Review * 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. Mole ratio -moles • • Mole moles • Molar mass moles grams • Molarity moles liters soln • Molar volume - moles liters gas Core step in all stoichiometry problems!! 4. Check answer.

A. Molar Volume at STP 1 mol of a gas=22. 4 L at STP Standard Temperature & 0°C and 1 atm Pressure

A. Molar Volume at STP LITERS OF GAS AT STP Molar Volume (22. 4 L/mol) MASS IN GRAMS Molar Mass (g/mol) MOLES 6. 02 1023 particles/mol NUMBER OF PARTICLES

B. Gas Stoichiometry of one Gas Liters of another Gas: • Avogadro’s 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? Ø start with ideal gas law • Looking for liters of gas? Ø start with stoichiometry conv b Liters

C. Gas Stoichiometry Problem – STP b How many grams of KCl. O 3 are req’d to produce 9. 00 L of O 2 at STP? 2 KCl. O 3 2 KCl + 3 O 2 ? g 9. 00 L O 2 1 mol 2 mol 122. 55 O 2 KCl. O 3 g KCl. O 3 22. 4 L 3 mol O 2 = 32. 8 g 1 mol KCl. O 3

D. Gas Stoichiometry Problem – Non-STP b What volume of CO 2 forms from 5. 25 g of Ca. CO 3 at 103 k. Pa & 25ºC? Ca. CO 3 5. 25 g Ca. O Looking for liters: Start with stoich and calculate moles of CO 2. 5. 25 g 1 mol Ca. CO 3 CO 2 + CO 2 ? L non. P = 103 k. Pa STP NEXT V=? n=? = 0. 0525 mol R = 8. 315 3 k. Pa/mol. K 100. 09 1 mol Plug this dmthe into CO 2 Ideal Gas Law for n to find liters g Ca. C T = 298 K

D. Gas Stoichiometry Problem – Non-STP b What volume of CO 2 forms from 5. 25 g of Ca. CO 3 at 103 k. Pa & 25ºC? GIVEN: WORK: P = 103 k. Pa PV = n. RT V=? (103 k. Pa)V =(0. 0525 mol)(8. 315 dm 3 k. Pa/ n = 0. 0525 mol T = 25°C = 298 K mol K) (298 K) R = 8. 315 3 dm 3 k. Pa/mol K V = 1. 26 dm CO 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 k. Pa & 21°C? 4 Al + GIVEN: P = 97. 3 k. Pa V = 15. 0 L n=? T = 21°C = 294 K R = 8. 315 3 O 2 15. 0 L non. STP WORK: 2 Al 2 O 3 ? g Given liters: Start with Ideal Gas Law and calculate moles of O 2. PV = n. RT (97. 3 k. Pa) (15. 0 L) = n (8. 315 dm 3 k. Pa/mol K) NEXT (294 K)

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 k. Pa & 21°C? 3 O 2 Use stoich to convert moles 15. 0 L of O to grams Al O non 0. 597 2 mol STP 101. 96 mol O 2 Al 2 O 3 g Al 2 O 3 4 Al 2 2 + 2 Al 2 O 3 ? g 3 3 mol O 2 = 40. 6 g 1 mol Al 2 O 3