1. 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

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

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

5. 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)

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?
P1 = 1300 torr
V1= X ml
T1 = 496K
P2 = 663 torr
V2 = ml
T2 = 592K

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

8. 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)

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?
P1 = X
V1= 2.7L
T1 = 466K
P2 = 1.01 atm
V2 = L
T2 = 605K

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

11. 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)

12. 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=
P= 1.00 atm
V=100L

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

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

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

16. 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 liter container in order to exert a pressure of atmospheres at 25 °C?
PV = nRT
T =298K
n = X mol R=
P= 2.00 atm
V=3.00L

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

19. ____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?

20. ____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

21. ____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

22. 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=
T= = 273K
(XL)(1.00 atm) =(23.8 mol)( )(273K)

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

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

25. 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)

26. 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=
T= = 473K

27. 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=
T= = 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 CH4
2 L H2O
= 21 L of H2O
1 L CH4
Now we can plug numbers into the equation.

29. 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=
T= = 473K
(21L)(1.00 atm) =(X mol)( )(473K)

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

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

32. 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.

33. 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=
T= = 273K

34. 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=
T= = 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. 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
= moles of O2
4 mole Fe
NOW!! Finally we can plug in mol of O2 in to equation

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

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

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

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

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

41. 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 = torr
P1=760 torr
V1= 500 ml
P2= torr
V2=X
(760)(500) = (738.9 )(x)
514 mL

42. 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 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 = kPa – 2.98 kPa
P1oxygen at 24°C = kPa
V1= 37.8 ml
T1= 297K
P2oxygen at STP= kPa
V2= X
T2= 273K
(99.42 kPa)(37.8 mL)(273 K)
(297 K)(101.3 kPa) =
(V2) =
34.1 mL

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