Entry Task: Dec 7th Block 1

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Presentation transcript:

Entry Task: Dec 7th Block 1 Collect Post Lab on Molar Mass of VL Lab Discuss three handouts Combination/Ideal ws Ideal and Stoich ws All Gas Law w.s Discuss Practice Test

Application of Combined Gas Law If a gas occupies a volume of 100 cm3 at a pressure of 101.3 kPa and 27C, what volume will the gas occupy at 120 kPa and 50C? P1 = 101.3 kPa V1= 100 cm3 T1 = 300K P2 = 120 kPa V2 = X T2 = 323K

Application of Combined Gas Law P1 = 101.3 kPa V1= 100 cm3 T1 = 300K P2 = 120 kPa V2 = X T2 = 323K (101.3) (100) = (120) (X cm3) 300 323

GET X by its self!! 323 300 = 91 cm3 (300)(120) (101.3) (100) = (120) (X cm3) 300 323 (101.3)(100)(323) = 91 cm3 (300)(120)

Combined Gas Law 2. A closed gas system initially has pressure and temperature of 1300 torr and 496 K with the volume unknown. If the same closed system has values of 663 torr, 8060 ml and 592K, what was the initial volume in ml? P1 = 1300 torr V1= X ml T1 = 496K P2 = 663 torr V2 = 8060 ml T2 = 592K

Application of Combined Gas Law P1 = 1300 torr V1= X ml T1 = 496K P2 = 663 torr V2 = 8060 ml T2 = 592K (1300 torr) (X ml) = (663 torr) (8060 ml) 496 K 592 K

GET X by its self!! 3440 ml (1300 torr)(592 K) 592 K 496 K (1300 torr) (X ml) = (663 torr) (8060 ml) 496 K 592 K (496 K)(693 torr)(8060 ml) 3440 ml (1300 torr)(592 K)

Combined Gas Law 3. A closed gas system initially has volume and temperature of 2.7 L and 466 K with the pressure unknown. If the same closed system has values of 1.01 atm, 4.70 L and 605 K, what was the initial pressure in atm? P1 = X V1= 2.7L T1 = 466K P2 = 1.01 atm V2 = 4.70 L T2 = 605K

Application of Combined Gas Law P1 = X V1= 2.7L T1 = 466K P2 = 1.01 atm V2 = 4.70 L T2 = 605K (X atm) (2.7L) = (1.01 atm) (4.70L) 466 K 605 K

GET X by its self!! 1.4 atm (2.7L)(605K) 605 K 466 K (X atm) (2.7L) = (1.01 atm) (4.70L) 466 K 605 K (466 K)(1.01atm)(4.70L) 1.4 atm (2.7L)(605K)

PV = nRT T =X n = 4.0 mol P= 1.00 atm R= 0.0821 V=100L 4. At what temperature (in Kelvin) would 4.0 moles of Hydrogen gas in a 100 liter container exert a pressure of 1.00 atmospheres? PV = nRT T =X n = 4.0 mol R= 0.0821 P= 1.00 atm V=100L

= 305K (1.00 atm)(100L) (0.0821 )(4.0 mol) (1.00 atm) (100 L) = (4.0 mol)(0.0821 )(XK) (1.00 atm)(100L) = 305K (0.0821 )(4.0 mol)

PV = nRT T =318K n = 0.5 mol P= X atm R= 0.0821 V=18L 5. An 18 liter container holds 16.00 grams of O2 at 45°C. What is the pressure (atm) of the container? PV = nRT T =318K n = 0.5 mol R= 0.0821 P= X atm V=18L

= 0.73 atm (0.5 mol)(0.0821 ) (318k) 18L (X atm) (18 L)

PV = nRT T =298K n = X mol P= 2.00 atm R= 0.0821 V=3.00L 6. How many moles of oxygen must be in a 3.00 liter container in order to exert a pressure of 2.00 atmospheres at 25 °C? PV = nRT T =298K n = X mol R= 0.0821 P= 2.00 atm V=3.00L

= 0.25 moles (2.00 atm)(3.00 L) (0.0821 )(298K) (2.00 atm) (3.00 L) = (X mol)(0.0821 )(298K) (2.00 atm)(3.00 L) = 0.25 moles (0.0821 )(298K)

____CaCO3(s)  ____CaO(s) + ____CO2(g) 7. Calcium carbonate forms limestone, one of the most common rocks on Earth. It also forms stalactites, stalagmites, and many other types of formations found in caves. When calcium carbonate is heated, it decomposes to form solid calcium oxide and carbon dioxide gas. ____CaCO3(s)  ____CaO(s) + ____CO2(g) How many liters of carbon dioxide will be produced at STP if 2.38 kg of calcium carbonate reacts completely?

____CaCO3(s)  ____CaO(s) + ____CO2(g) How many liters of carbon dioxide will be produced at STP if 2.38 kg of calcium carbonate reacts completely? Balance equation- The equation shows that there is a 1:1 ratio between CO2 and CaCO3

____CaCO3(s)  ____CaO(s) + ____CO2(g) How many liters of carbon dioxide will be produced at STP if 2.38 kg of calcium carbonate reacts completely? Now we need to change 2.38 kg to grams, then grams to moles. 2.38 kg of CaCO3 1 mole CaCO3 1000g 1 kg 100. g CaCO3 = 23.8 mol CaCO3 Since 23.8 mol of CaCO3 and there is a 1:1 ratio to CO2, then there is 23.8 moles of CO2

PV = nRT VCO2 = X L n = 23.8 mol of CO2 P = 1.00 atm R= 0.0821 How many liters of carbon dioxide will be produced at STP if 2.38 kg of calcium carbonate reacts completely? PV = nRT VCO2 = X L P = 1.00 atm n = 23.8 mol of CO2 R= 0.0821 T= 0.00 + 273 = 273K (XL)(1.00 atm) =(23.8 mol)(0.0821 )(273K)

PV = nRT (23.8 mol)(0.0821 )(273K) X L = (1.00 atm) (XL)(1.00 atm) =(23.8 mol)(0.0821 )(273K) (23.8 mol)(0.0821 )(273K) X L = (1.00 atm)

DO the MATH (23.8)(0.0821L)(273) X L = (1.00 ) 533.4 = 533 L 1.0

8. Determine how many moles of water vapor will be produced at 1 8. Determine how many moles of water vapor will be produced at 1.00 atm and 200°C by the complete combustion of 10.5 L of methane gas (CH4). 2 2 __CH4 (g) + __O2 (g)  __CO2(g) + __H2O(g)

PV = nRT 2 2 VCH4 = 10.5 L n = X mol of H2O P = 1.00 atm R= 0.0821 8. Determine how many moles of water vapor will be produced at 1.00 atm and 200°C by the complete combustion of 10.5 L of methane gas (CH4). 2 2 __CH4 (g) + __O2 (g)  __CO2(g) + __H2O(g) PV = nRT VCH4 = 10.5 L P = 1.00 atm n = X mol of H2O R= 0.0821 T= 200 + 273 = 473K

PV = nRT 2 2 VCH4 = 10.5 L n = X mol of H2O P = 1.00 atm R= 0.0821 __CH4 (g) + __O2 (g)  __CO2(g) + __H2O(g) PV = nRT VCH4 = 10.5 L P = 1.00 atm n = X mol of H2O R= 0.0821 T= 200 + 273 = 473K We are given 10.5 liters of CH4, we need to find out how many liters of water to plug into equation.

10.5L of CH4 2 L H2O = 21 L of H2O 1 L CH4 Now we can plug numbers into the equation.

PV = nRT VH2O = 21 L n = X mol of H2O P = 1.00 atm R= 0.0821 8. Determine how many moles of water vapor will be produced at 1.00 atm and 200°C by the complete combustion of 10.5 L of methane gas (CH4). PV = nRT VH2O = 21 L P = 1.00 atm n = X mol of H2O R= 0.0821 T= 200 + 273 = 473K (21L)(1.00 atm) =(X mol)(0.0821 )(473K)

PV = nRT (21 L)(1.00 atm) = X mol (0.0821 )(473K)

DO the MATH (21 )(1.00) = X mol (0.0821 mol)(473) 21 = 0.541 mol 38.83

4 3 2 ____Fe(s) + ___O2(g)  ___Fe2O3(s) 9. When iron rusts, it undergoes a reaction with oxygen to form iron (III) oxide. 4 3 2 ____Fe(s) + ___O2(g)  ___Fe2O3(s) Calculate the volume of oxygen gas at STP that is required to completely react with 52.0 g of iron.

PV = nRT 4 3 2 VO2 = X L n = 52.0g of Fe P = 1.00 atm R= 0.0821 9. When iron rusts, it undergoes a reaction with oxygen to form iron (III) oxide. 4 3 2 ____Fe(s) + ___O2(g)  ___Fe2O3(s) Calculate the volume of oxygen gas at STP that is required to completely react with 52.0 g of iron. PV = nRT VO2 = X L P = 1.00 atm n = 52.0g of Fe R= 0.0821 T= 0.00 + 273 = 273K

PV = nRT 4 3 2 VO2 = X L n = 52.0g of Fe P = 1.00 atm R= 0.0821 ____Fe(s) + ___O2(g)  ___Fe2O3(s) Calculate the volume of oxygen gas at STP that is required to completely react with 52.0 g of iron. PV = nRT VO2 = X L P = 1.00 atm n = 52.0g of Fe R= 0.0821 T= 0.00 + 273 = 273K We are given 52.0 grams of Fe, we need to find out how many mole this will be. From this, we can compare mole ratio to get # moles for oxygen.

4 3 2 ____Fe(s) + ___O2(g)  ___Fe2O3(s) 52.0 g of Fe 1 mole Fe = 0.93 moles of Fe 55.85 g Fe 4 3 2 ____Fe(s) + ___O2(g)  ___Fe2O3(s) From this we can get mole to mole ratio to get O2 0.93 mol Fe 3 moleO2 = 0.698 moles of O2 4 mole Fe NOW!! Finally we can plug in 0.698 mol of O2 in to equation

PV = nRT VO2 = X L n = 0.698 mol O2 P = 1.00 atm R= 0.0821 T= 0.00 + 273 = 273K (XL)(1.00 atm) =(0.698 mol)(0.0821 )(273K)

PV = nRT (0.698 mol)(0.0821 )(273K) X L = (1.00 atm) (XL)(1.00 atm) =(0.698 mol)(0.0821 )(273K) (0.698 mol)(0.0821 )(273K) X L = (1.00 atm)

DO the MATH (0.698)(0.0821 L)(273) X L = (1.00) 15.6 = 15.6 L 1.00

Partial Pressure-Collecting gas over water Nitrogen gas is collected at 20°C and a total ambient pressure of 735.8 torr using the method of water displacement. What is the partial pressure of dry nitrogen? PTOTAL = 735.8 torr PWATER= 17.5 torr (20 °C) PNitrogen = ? PTOTAL = Pnitrogen + PWATER PNitrogen = PTOTAL – PWATER = 735.8 torr – 17.5 torr PNitrogen = 718.3 torr

Partial Pressure-Collecting gas over water Hydrogen gas is collected over water at a total pressure of 714.4 mm Hg. The volume of hydrogen collected is 30 mL at 25°C. What is the partial pressure of hydrogen gas? PHydrogen = PTOTAL – PWATER = 714.4 mmHg– 23.7 mmHg Phydrogen = 690.7 mmHg (690.7 torr) P1=714.4 torr V1= 30 ml P2= 690.7 torr V2=X (714.4)(30) = (690.7)(x) 31.0 mL

Partial Pressure-Collecting gas over water A 500 mL sample of oxygen was collected over water at 23°C and 760 torr pressure. What volume will the dry oxygen occupy at 23°C and 760 torr? The vapor pressure of water at 23°C is 21.1 torr. Poxygen = PTOTAL – PWATER = 760 torr – 21.1 torr Poxygen = 738.9 torr P1=760 torr V1= 500 ml P2= 738.9 torr V2=X (760)(500) = (738.9 )(x) 514 mL

Partial Pressure-Collecting gas over water 37.8 mL of O2 is collected by the downward displacement of water at 24°C and an atmospheric pressure of 102.4 kPa. What is the volume of dry oxygen measured at STP (noticed we changed temp and pressure) (Patm=101.3 kPa)? Use P1V1/T1 = P2V2/T2 Poxygen = PTOTAL – PWATER = 102.4 kPa – 2.98 kPa P1oxygen at 24°C = 99.42 kPa V1= 37.8 ml T1= 297K P2oxygen at STP= 101.3 kPa V2= X T2= 273K (99.42 kPa)(37.8 mL)(273 K) (297 K)(101.3 kPa) = (V2) = 34.1 mL

Collect gas over water Lab HW: Post Lab Questions AND Ch. 10 sec. 7-10 reading notes