Entry Task: Dec 7th Block 1Entry 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

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 27C, what volume will the gas occupy at 120 kPa and 50C?

P1 = 101.3 kPa

V1= 100 cm3

T1 = 300K

P2 = 120 kPa

V2 = X

T2 = 323K

Application of Combined Gas Law

(101.3) (100)

300 323

= (120) (X cm3)

P1 = 101.3 kPa

V1= 100 cm3

T1 = 300K

P2 = 120 kPa

V2 = X

T2 = 323K

GET X by its self!!

(101.3)(100)(323)

(300)(120)

(101.3) (100)

300 323

= (120) (X cm3)

= 91 cm3

Combined Gas Law2. 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

(1300 torr) (X ml)

496 K 592 K

= (663 torr) (8060 ml)

P1 = 1300 torr

V1= X ml

T1 = 496K

P2 = 663 torr

V2 = 8060 ml

T2 = 592K

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

Combined Gas Law3. 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

(X atm) (2.7L)

466 K 605 K

= (1.01 atm) (4.70L)

P1 = X

V1= 2.7L

T1 = 466K

P2 = 1.01 atm

V2 = 4.70 L

T2 = 605K

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)

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 atmV=100L

T =Xn = 4.0 molR= 0.0821

PV = nRT

(0.0821 )(4.0 mol)

(1.00 atm)(100L)= 305K

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

5. An 18 liter container holds 16.00 grams of O2 at 45°C. What is the pressure (atm) of the container?

P= X atmV=18L

T =318Kn = 0.5 molR= 0.0821

PV = nRT

18L

(0.5 mol)(0.0821 ) (318k)= 0.73 atm

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

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 atmV=3.00L

T =298Kn = X molR= 0.0821

PV = nRT

(0.0821 )(298K)

(2.00 atm)(3.00 L)= 0.25 moles

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

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

2.38 kg of CaCO3

1 kg

1000g

= 23.8 mol 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?

100. g CaCO3

1 mole CaCO3

Now we need to change 2.38 kg to grams, then grams to moles.

Since 23.8 mol of CaCO3 and there is a 1:1 ratio to CO2, then there is 23.8 moles of CO2

PV = nRTVCO2 = X L

P = 1.00 atmn = 23.8 mol of CO2

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?

PV = nRT

(23.8 mol)(0.0821 )(273K)

(1.00 atm)X L =

(XL)(1.00 atm) =(23.8 mol)(0.0821 )(273K)

DO the MATH

533.41.0 = 533 L

(23.8)(0.0821L)(273)

(1.00 )X L =

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).

__CH4 (g) + __O2 (g) __CO2(g) + __H2O(g)22

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).

__CH4 (g) + __O2 (g) __CO2(g) + __H2O(g)22

PV = nRTVCH4 = 10.5 L

P = 1.00 atmn = X mol of H2O

R= 0.0821 T= 200 + 273 = 473K

__CH4 (g) + __O2 (g) __CO2(g) + __H2O(g)22

PV = nRTVCH4 = 10.5 L

P = 1.00 atmn = 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 = 21 L of H2O2 L H2O

1 L CH4

Now we can plug numbers into the equation.

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 = nRTVH2O = 21 L

P = 1.00 atmn = 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)

(0.0821 )(473K)= X mol

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

DO the MATH

2138.83

= 0.541 mol

(21 )(1.00)

(0.0821 mol)(473)= X mol

9. When iron rusts, it undergoes a reaction with oxygen to form iron (III) oxide.

____Fe(s) + ___O2(g) ___Fe2O3(s)23

Calculate the volume of oxygen gas at STP that is required to completely react with 52.0 g of iron.

4

9. When iron rusts, it undergoes a reaction with oxygen to form iron (III) oxide.

____Fe(s) + ___O2(g) ___Fe2O3(s)23Calculate the volume of oxygen gas at STP that is

required to completely react with 52.0 g of iron.

4

PV = nRTVO2 = X L

P = 1.00 atmn = 52.0g of FeR= 0.0821 T= 0.00 + 273 = 273K

____Fe(s) + ___O2(g) ___Fe2O3(s)23Calculate the volume of oxygen gas at STP that is

required to completely react with 52.0 g of iron.

4

PV = nRTVO2 = X L

P = 1.00 atmn = 52.0g of FeR= 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.

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 O2

____Fe(s) + ___O2(g) ___Fe2O3(s)234

0.93 mol Fe= 0.698 moles of O2

3 moleO2

4 mole Fe

NOW!! Finally we can plug in 0.698 mol of O2 in to equation

PV = nRTVO2 = X L

P = 1.00 atmn = 0.698 mol O2

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)

(1.00 atm)X L =

(XL)(1.00 atm) =(0.698 mol)(0.0821 )(273K)

DO the MATH

15.61.00

= 15.6 L

(0.698)(0.0821 L)(273)

(1.00)X L =

Partial Pressure-Collecting gas over waterNitrogen 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 torrPWATER= 17.5 torr (20 °C)PNitrogen = ?

PTOTAL = Pnitrogen + PWATER

PNitrogen = PTOTAL – PWATER

= 735.8 torr – 17.5 torr PNitrogen = 718.3 torr

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 torrV1= 30 mlP2= 690.7 torrV2=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 torrV1= 500 mlP2= 738.9 torrV2=X

(760)(500) = (738.9 )(x)514 mL

Partial Pressure-Collecting gas over water

Partial Pressure-Collecting gas over water37.8 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 kPaV1= 37.8 mlT1= 297K

P2oxygen at STP= 101.3 kPaV2= XT2= 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