Download presentation

Presentation is loading. Please wait.

Published byTianna Holstead Modified about 1 year ago

1
Entry Task: Dec 7 th 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 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

3
Application of Combined Gas Law If a gas occupies a volume of 100 cm 3 at a pressure of 101.3 kPa and 27 C, what volume will the gas occupy at 120 kPa and 50 C? P 1 = 101.3 kPa V 1 = 100 cm 3 T 1 = 300K P 2 = 120 kPa V 2 = X T 2 = 323K

4
Application of Combined Gas Law (101.3) (100) 300 323 = (120) (X cm 3 ) P 1 = 101.3 kPa V 1 = 100 cm 3 T 1 = 300K P 2 = 120 kPa V 2 = X T 2 = 323K

5
GET X by its self!! (101.3)(100)(323) (300)(120) (101.3) (100) 300 323 = (120) (X cm 3 ) = 91 cm 3

6
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? P 1 = 1300 torr V 1 = X ml T 1 = 496K P 2 = 663 torr V 2 = 8060 ml T 2 = 592K

7
Application of Combined Gas Law (1300 torr) (X ml) 496 K 592 K = (663 torr) (8060 ml) P 1 = 1300 torr V 1 = X ml T 1 = 496K P 2 = 663 torr V 2 = 8060 ml T 2 = 592K

8
GET X by its self!! (496 K)(693 torr)(8060 ml) (1300 torr)(592 K) (1300 torr) (X ml) 496 K 592 K = (663 torr) (8060 ml) 3440 ml

9
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? P 1 = X V 1 = 2.7L T 1 = 466K P 2 = 1.01 atm V 2 = 4.70 L T 2 = 605K

10
Application of Combined Gas Law (X atm) (2.7L) 466 K 605 K = (1.01 atm) (4.70L) P 1 = X V 1 = 2.7L T 1 = 466K P 2 = 1.01 atm V 2 = 4.70 L T 2 = 605K

11
GET X by its self!! (466 K)(1.01atm)(4.70L) (2.7L)(605K) 1.4 atm (X atm) (2.7L) 466 K 605 K = (1.01 atm) (4.70L)

12
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? P= 1.00 atm V=100L T =X n = 4.0 mol R= 0.0821 PV = nRT

13
(0.0821 )(4.0 mol) (1.00 atm)(100L) = 305K (1.00 atm) (100 L) = (4.0 mol)(0.0821 )(XK)

14
5. An 18 liter container holds 16.00 grams of O 2 at 45°C. What is the pressure (atm) of the container? P= X atm V=18L T =318K n = 0.5 mol R= 0.0821 PV = nRT

15
18L (0.5 mol)(0.0821 ) (318k) = 0.73 atm (X atm) (18 L) = (0.5 mol)(0.0821 )(318K)

16
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? P= 2.00 atm V=3.00L T =298K n = X mol R= 0.0821 PV = nRT

17
(0.0821 )(298K) (2.00 atm)(3.00 L) = 0.25 moles (2.00 atm) (3.00 L) = (X mol)(0.0821 )(298K)

19
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. ____CaCO 3 (s) ____CaO(s) + ____CO 2 (g) How many liters of carbon dioxide will be produced at STP if 2.38 kg of calcium carbonate reacts completely?

20
____CaCO 3 (s) ____CaO(s) + ____CO 2 (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 CO 2 and CaCO 3

21
2.38 kg of CaCO 3 1 kg 1000g = 23.8 mol CaCO 3 ____CaCO 3 (s) ____CaO(s) + ____CO 2 (g) How many liters of carbon dioxide will be produced at STP if 2.38 kg of calcium carbonate reacts completely? 100. g CaCO 3 1 mole CaCO 3 Now we need to change 2.38 kg to grams, then grams to moles. Since 23.8 mol of CaCO 3 and there is a 1:1 ratio to CO 2, then there is 23.8 moles of CO 2

22
PV = nRT V CO2 = X L P = 1.00 atm n = 23.8 mol of CO 2 R= 0.0821 T= 0.00 + 273 = 273K (XL)(1.00 atm) =(23.8 mol)(0.0821 )(273K) How many liters of carbon dioxide will be produced at STP if 2.38 kg of calcium carbonate reacts completely?

23
PV = nRT (23.8 mol)(0.0821 )(273K) (1.00 atm) X L = (XL)(1.00 atm) =(23.8 mol)(0.0821 )(273K)

24
DO the MATH 533.4 1.0 = 533 L (23.8)(0.0821L)(273) (1.00 ) X L =

25
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 (CH 4 ). __CH 4 (g) + __O 2 (g) __CO 2 (g) + __H 2 O(g) 22

26
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 (CH 4 ). __CH 4 (g) + __O 2 (g) __CO 2 (g) + __H 2 O(g) 22 PV = nRT V CH4 = 10.5 L P = 1.00 atm n = X mol of H 2 O R= 0.0821 T= 200 + 273 = 473K

27
__CH 4 (g) + __O 2 (g) __CO 2 (g) + __H 2 O(g) 22 PV = nRT V CH4 = 10.5 L P = 1.00 atm n = X mol of H 2 O 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.

28
10.5L of CH 4 = 21 L of H 2 O 2 L H 2 O 1 L CH 4 Now we can plug numbers into the equation.

29
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 (CH 4 ). PV = nRT V H2O = 21 L P = 1.00 atm n = X mol of H 2 O R= 0.0821 T= 200 + 273 = 473K (21L)(1.00 atm) =(X mol)(0.0821 )(473K)

30
PV = nRT (21 L)(1.00 atm) (0.0821 )(473K) = X mol (21L)(1.00 atm) =(X mol)(0.0821 )(473K)

31
DO the MATH 21 38.83 = 0.541 mol (21 )(1.00) (0.0821 mol)(473) = X mol

32
9. When iron rusts, it undergoes a reaction with oxygen to form iron (III) oxide. ____Fe(s) + ___O 2 (g) ___Fe 2 O 3 (s) 23 Calculate the volume of oxygen gas at STP that is required to completely react with 52.0 g of iron. 4

33
9. When iron rusts, it undergoes a reaction with oxygen to form iron (III) oxide. ____Fe(s) + ___O 2 (g) ___Fe 2 O 3 (s) 23 Calculate the volume of oxygen gas at STP that is required to completely react with 52.0 g of iron. 4 PV = nRT V O2 = X L P = 1.00 atm n = 52.0g of Fe R= 0.0821 T= 0.00 + 273 = 273K

34
____Fe(s) + ___O 2 (g) ___Fe 2 O 3 (s) 23 Calculate the volume of oxygen gas at STP that is required to completely react with 52.0 g of iron. 4 PV = nRT V O2 = 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.

35
52.0 g of Fe = 0.93 moles of Fe 1 mole Fe 55.85 g Fe From this we can get mole to mole ratio to get O 2 ____Fe(s) + ___O 2 (g) ___Fe 2 O 3 (s) 234 0.93 mol Fe = 0.698 moles of O 2 3 moleO 2 4 mole Fe NOW!! Finally we can plug in 0.698 mol of O 2 in to equation

36
PV = nRT V O2 = X L P = 1.00 atm n = 0.698 mol O 2 R= 0.0821 T= 0.00 + 273 = 273K (XL)(1.00 atm) =(0.698 mol)(0.0821 )(273K)

37
PV = nRT (0.698 mol)(0.0821 )(273K) (1.00 atm) X L = (XL)(1.00 atm) =(0.698 mol)(0.0821 )(273K)

38
DO the MATH 15.6 1.00 = 15.6 L (0.698)(0.0821 L)(273) (1.00) X L =

39
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? P TOTAL = 735.8 torr P WATER = 17.5 torr (20 °C) P Nitrogen = ? P TOTAL = P nitrogen + P WATER P Nitrogen = P TOTAL – P WATER = 735.8 torr – 17.5 torr P Nitrogen = 718.3 torr

40
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? P Hydrogen = P TOTAL – P WATER = 714.4 mmHg– 23.7 mmHg P hydrogen = 690.7 mmHg (690.7 torr) P 1 =714.4 torr V 1 = 30 ml P 2 = 690.7 torr V 2 =X (714.4)(30) = (690.7)(x) 31.0 mL Partial Pressure-Collecting gas over water

41
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. P oxygen = P TOTAL – P WATER = 760 torr – 21.1 torr P oxygen = 738.9 torr P 1 =760 torr V 1 = 500 ml P 2 = 738.9 torr V 2 =X (760)(500) = (738.9 )(x) 514 mL Partial Pressure-Collecting gas over water

42
37.8 mL of O 2 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) (P atm =101.3 kPa)? Use P 1 V 1 /T 1 = P 2 V 2 /T 2 P oxygen = P TOTAL – P WATER = 102.4 kPa – 2.98 kPa P 1oxygen at 24°C = 99.42 kPa V 1 = 37.8 ml T 1 = 297K P 2oxygen at STP = 101.3 kPa V 2 = X T 2 = 273K (99.42 kPa)(37.8 mL)(273 K) (297 K)(101.3 kPa) =(V 2 ) =34.1 mL

43
Collect gas over water Lab HW: Post Lab Questions AND Ch. 10 sec. 7-10 reading notes

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google