law of gases
DESCRIPTION
Dalton's Law, Avogadro's Law, Molar Volume, and Ideal Gas LawTRANSCRIPT
![Page 1: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/1.jpg)
Project in Chemistry
SY: 2012-2013
Submitted by: Axl Merk Mindajao
Mary Rose Baccay
III-OBB
Submitted to: Mrs. Panlasiqui
![Page 2: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/2.jpg)
Recall 1. If the pressure exerted on a gas is
tripled, what will happen to the volume (assuming the temperature and amount of gas remains constant)?
A. x2 C. x 1/2
B. X3 D. x 1/3
![Page 3: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/3.jpg)
2. Carbon dioxide occupies a 2.54 L container at STP. What will be the volume when the pressure is 150 KPa and 26 C?
A. 1.89 L C. 0.163 L
B. 42300 L D. 14.1 L
![Page 4: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/4.jpg)
3. In which of the following sets of conditions would a gas most act as a real gas?
A. high temperature and low pressure
B. high temperature and high pressure
C. low temperature and high pressure
D. low temperature and low pressure
![Page 5: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/5.jpg)
4. Which of the following terms is used to describe the average kinetic energy of the molecules in a sample?
A. manometer C. pressure
B. temperature D. volume
![Page 6: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/6.jpg)
5. Which of the following is not a standard pressure measurement?
A. 101.3 kPa C. 760 mm Hg
B. 1 atm D. 273 K
![Page 7: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/7.jpg)
6. The volume of a gas is increased from 150.0 mL to 350.0 mL by heating it. If the original temperature of the gas was 25.0 °C, what is the final temperature?
A. -146 °C C. 695 °C
B. 10.7 °C D. 150 °C
![Page 8: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/8.jpg)
7. What is the name of the phase change that exists when a solid turns into a liquid?
A. depositon C. sublimation
B. melting D. freezing
![Page 9: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/9.jpg)
8. A balloon is inflated to a volume of 130 ml at a pressure of 690 mmHg. If the pressure is increased to 1000 mmHg, what will the new volume be?
A. 188 ml C. 89.7 ml
B. 98.8 ml D. 40300 ml
![Page 10: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/10.jpg)
9. What type of relationship is shown between temperature and pressure?
A. direct C. exponential
B. inverse D. logarithmic
![Page 11: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/11.jpg)
10. How many mm Hg is equal to 121 kPa?
A. 1.19 C. 636 B. 16.1 D. 908
![Page 12: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/12.jpg)
Answers1.d 6. c2.a 7. b3.c 8. c4.b 9. a5.d 10. d
![Page 13: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/13.jpg)
TitleDalton’s Law of Partial
Pressure
![Page 14: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/14.jpg)
Objectives To demonstrate the ability to use
Dalton’s Law of Partial Pressure in calculations.
· To calculate the total pressure of a mixture of gases or the partial pressure of a gas in a mixture of gases.
· To employ Dalton’s Law of Partial Pressure to predict the pressure of a gas mixture.
![Page 15: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/15.jpg)
Unlocking of Words• Pressure- the force exerted by gas
particles that hit the walls of a container; a force applied per unit area.
• Partial pressure- the individual pressure of each gas in a mixture.
• Water vapor- gaseous water found in the air that is below the boiling point of water.
![Page 16: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/16.jpg)
Lesson• The English chemist John Dalton
investigated pressures in mixture of gases. Dalton’s Law of Partial Pressure states that: “At constant volume and temperature, the total pressure in a mixture of gases is equal to the sum of the partial pressures of the component gases.”
![Page 17: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/17.jpg)
• In equation form, PT=P1+P2+....where
PT= total pressure
P1= partial pressure of gas 1
P2= partial pressure of gas 2
![Page 18: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/18.jpg)
ActivityProblem 1: Oxygen gas occupies
500 ml at 20°C and 760 mmHg. What volume will it occupy if it is collected over water at 30°C and 750 mmHg? (Water vapor pressure at 30°C31.8 mmHg)
![Page 19: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/19.jpg)
• Solution: Step 1 Given: P1= 760 mmHg V1= 500 ml T1= 20°+273 = 293 K P2 = 750 mmHg-31.8 mmHg = 718.2 mmHg T2 = 30°C+273 = 303 K
![Page 20: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/20.jpg)
Step 2: The combined gas formulaP1V1= P2V2 V2= P1 T2 T1 T2 P2 T1
Step 3:V2= 500 ml 760 mmHg 303 K 718.2 mmHg 93 K V2= 547 ml
![Page 21: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/21.jpg)
Problem 2: Calculate the mass of 400. mL of carbon dioxide collected over water at 30° C and 749 mm Hg.
Solution:PT = Pgas + Pwater= 749 mmHg R = 0.0821 L·atm/mol·K Pgas = 749 mmHg–31.8 mmHg = 717 mmHgV = 400.0 L
![Page 22: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/22.jpg)
T = 30° C + 273 = 303 K PV = nRT n = PV/RT n = 717 mm x 1 atm/760 mm x 400.0 mL x
1L/10 ³mL/(0.0821 L·atm/mol·K x 303 K) n = 0.0152 mol CO2
n = 0.0152 mol CO2 x 44.01 g CO2/1 mol CO2 = 0.669 g CO2
![Page 23: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/23.jpg)
Exercise1. Determine the total pressure of a
gas mixture that contains oxygen, nitrogen, and helium if the partial pressures of the gases are: PO2 = 20.2 kPa
PN2 = 46.7 kPa
PHe = 26.7 kPa
![Page 24: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/24.jpg)
2. A 10 L flask at 298 K contains a
gaseous mixture of CO and CO2 at a total pressure of 1520 mm of Hg. If 0.20 mole of CO is present, find the partial pressure of CO and that of CO2?
![Page 25: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/25.jpg)
• Solution: 1. Ptotal = PO2 + PN2 + PHe
Ptotal = 20.2 + 46.7 + 26.7Ptotal = 93.6 kPa
2. PCO + PCO2 = P = 1520 mm of Hg
![Page 26: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/26.jpg)
= 0.49 atm.Partial pressure of CO2, pCO2= P - (pCO)
= 2.0 - 0.49 = 1147.6 mm of Hg
![Page 27: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/27.jpg)
TitleAvogadro’s Law
![Page 28: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/28.jpg)
Objectives• Determine how the amount of gas in
a fixed volume at a fixed pressure and temperature depends upon the identity of the gas.• To do calculations involving
Avogadro’s Law.
![Page 29: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/29.jpg)
Unlocking of Words• Moles- another name for grams.• Temperature- a measure of the
average kinetic energy of the particles of a substance.• Volume- the amount of space
occupied by something or within a container.
![Page 30: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/30.jpg)
Lesson• Amedeo Avogadro stated that
the volume of a gas is directly related to its number of moles when temperature and pressure remain unchanged.
![Page 31: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/31.jpg)
• If the moles of a gas are doubled, then the volume will double as long as the pressure and temperature remain the same.• To illustrate these two conditions,
you may write:V1=V2 n1 n2
![Page 32: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/32.jpg)
ActivityProblem 1: A balloon containing 2
moles of helium has a volume of 880 ml. What is the new volume after 4 more moles of helium are added to the balloon at the same temperature and pressure?
![Page 33: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/33.jpg)
Solution: Step 1Initial condition Final conditionV1= 880 ml V2=?n1= 2 moles n2= 6 molesStep 2V2= 880 ml x 6 moles =2640 ml
2 moles
![Page 34: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/34.jpg)
Problem 2: What volume of O2 measured at 1.86
x 106 Pa and 375 K, is needed to react completely with 1.78 L of H2,
measured at the same pressure and temperature, to give H2O?
![Page 35: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/35.jpg)
Solution:2H₂ O₂ → 2H₂O
The volume of O₂ needed is1.78 L H₂ (1L O₂ · 2L H₂)
= 0.890 L O₂
![Page 36: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/36.jpg)
Exercise1. How many molecules of O2 are
present in 1.00L of O2 at STP?
2. Calculate the number of moles of ammonia gas, NH3, in a volume of 80 L of the gas measured at STP.
![Page 37: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/37.jpg)
Solution: 1. ___1L STP____ = 0.0446 mol O2
22.4 L mol STP ⁻ⁱ
(6.02 × 1023 molecules mol-1)(0.0446 mol) = 2.69 × 1022molecules
![Page 38: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/38.jpg)
2. V= number of moles × 22.414 L/molno. of moles= volume of gas
22.414 = 80__
22.414 = 3.569 L
![Page 39: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/39.jpg)
Title
Molar Volume
![Page 40: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/40.jpg)
Objectives• To experimentally determine the
volume of one mole of a gas at standard temperature and pressure.• To experimentally determine the
value of the gas constant, R.• To calculate the standard molar
volume of a gas from accumulated data.
![Page 41: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/41.jpg)
Unlocking of Words• Molar mass- the mass of one mole
of an element or compound equal to the atomic or formula weight.• Molar volume- the volume in liters
of one mole of a gas at STP.• STP- the standard gas conditions of 1
atm of pressure and 273 K.
![Page 42: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/42.jpg)
Lesson• At STP conditions (1atm and 273K),
one mole of a gas occupies 22.4 liters. • Molar volume is an extension of
Avogadro’s Law, w/c state that,”At any given temp. and pressure equal volume of any gases would contain equal no. of moles.”
![Page 43: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/43.jpg)
ActivityProblem 1: What is the volume of 56 g N ₂gas at STP?Solution:V= 56g N x ₂ 1 mole N2 x 22.4 L
N₂ 28 g N 1 mole ₂ N₂ =44.8 L N ₂
![Page 44: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/44.jpg)
Problem 2: How many liters of 0.250 moles of HCl
will occupy at STP?Solution:(x/0.250 mol)=(22.414L/1 mol)0.250 mol x 22.414 L molX= 5.60 L
![Page 45: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/45.jpg)
Exercise1. Find out the volume contained by
6.8 g of ammonia at STP?2. A balloon contains 0.5 moles of pure
helium gas at standard temperature and pressure. What is the volume of the balloon?
![Page 46: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/46.jpg)
Solution:1. Gram molecular mass of NH =17 g₃Molar volume= 22.4 liters17g : 22.4 L6.8 g : xx= (6.8 X 22.4)/17x= 8.96 Liters
![Page 47: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/47.jpg)
2. Extract the data from the question: n(He) = 0.5 mol Vm = 22.71 L mol-1 (at STP 1 mole of gas occupies 22.71 L) V(He) = ? L
Write the equation: V(He) = n(He) x Vm
Substitute in the values and solve: V(He) = 0.5 x 22.71 = 11.4 L
![Page 48: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/48.jpg)
Title
Ideal Gas Law
![Page 49: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/49.jpg)
Objectives• Demonstrate the ability to use the
ideal gas content to a basic calculation and those that involve density and molecular mass.• Calculate the amount of gas at any
specified conditions of pressure, volume and temperature.
![Page 50: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/50.jpg)
Unlocking of Words
• Ideal gas- an imaginary gas whose behavior is described by the gas laws.• Ideal gas law- a law that combines
the 4 meausrable properties of a gas in the equationPV= nRT
![Page 51: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/51.jpg)
Lesson• When the temperature, pressure and
volume of a gas areknown, ideal gas equation can be use.• The ideal gas equation, PV=nRT,
illustrates a direct relationships between volume, temperature and the no. of moles of a gas.
![Page 52: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/52.jpg)
• Volume and pressure are inversely related.
P= pressure in atomV= volume in litern= no. of moles of gasT= kelvin temperatureR= 0.082 L atm/Kmol
![Page 53: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/53.jpg)
ActivityProblem 1: How many molecules are
there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mmHg?
Solution:P = 1.00 x 10-6 mmHg T = 0.0° C + 273 = 273 K V = 985 mL R = 0.0821 L·atm/mol·K
![Page 54: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/54.jpg)
PV = nRT n = PV/RT n = 1.00 x 10-6 mm x 1 atm/760 mm x
985 mL x 1 L/103 mL/ (0.0821 L·atm/mol·K x 273 K)
= 5.78 x 10-11 moles N₂ n= 5.78 x 10-11 moles N₂ x 6.02 x 1023
N₂ molecules/1 mol N₂ = 3.48 x 1013 N₂ molecules
![Page 55: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/55.jpg)
Problem 2: Calculate the mass of 15.0 L of NH3 at 27° C and 900. mm Hg.
Solution:P = 900. mm Hg T = 27° C + 273 = 300 K V = 15.0 L R = 0.0821 L·atm/mol·K
![Page 56: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/56.jpg)
PV = nRT n = PV/RT n = 900. mm x 1 atm/760 mm x 15.0 L/(0.0821 L·atm/mol·K x 300 K) n = 0.721 moles NH3 x 17.04 g NH3/1 mol NH3 = 12.3 g NH3
![Page 57: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/57.jpg)
AssessmentI. Choose the best answer1. At the water’s surface, the
pressure on your body due to the mass of air around you is about
a. 760 KPa c. 100 mmb. 101.3 KPa d. 760 mm
![Page 58: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/58.jpg)
2. If the volume of mole of gas molecules remains constant, lowering the temperature will make the pressure
a. increaseb. increase then decrease c. decreased. decrease then increase
![Page 59: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/59.jpg)
3. If the volume available to the gas is increased, the pressure exerted by one mole of gas molecules will
a. increaseb. increase then decrease c. decreased. decrease then increase
![Page 60: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/60.jpg)
4. How many moles of O are ₂present in 44.8 L of O at STP?₂
a. 1.2 molesb. 1.4 molesc. 2.0 molesd. 2.8 moles
![Page 61: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/61.jpg)
5. What pressure must be applied to 225 ml of gas at 1 atm to reduce its volume to 100 ml?
a. 0.44 atmb. 2.25 atmc. 22500 atmd. 1250 atm
![Page 62: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/62.jpg)
Answer1. c2. a3. b4. d5. c
![Page 63: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/63.jpg)
II. Problem solving6. Calculate the density in g/L of
478 mL of krypton at 47° C and 671 mm Hg.
7. Calculate the mass of 400. mL of carbon dioxide collected over water at 30.° C and 749 mm Hg.
![Page 64: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/64.jpg)
8. Find out the volume contained by 6.8 g of ammonia at STP. (N=14, H=1)
9. What volume of hydrogen will react with 22.4 liters of oxygen to form water?
10. Find the volume from the 0.250 moles gas at 200kpa and 300K temperature.
![Page 65: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/65.jpg)
Solution⑥
P = 671 mm Hg T = 47° C + 273 = 320. K
V = 478 mL R = 0.0821 L·atm/mol·K
![Page 66: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/66.jpg)
PV = nRT n = m/MM D = m/V = P x MM/R x T D = 671 mm x 1 atm/760 mm x
83.80 g/mol/(0.0821 L·atm/mol·K x 320. K)
= 2.82 g/L
![Page 67: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/67.jpg)
⑦PT = Pgas + Pwater = 749 mm Hg R = 0.0821 L·atm/mol·K Pgas = 749 mm Hg – 31.8 mm Hg = 717 mm Hg V = 400.0 L T = 30.° C + 273 = 303 K
![Page 68: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/68.jpg)
n = PV/RT n = 717 mm x 1 atm/760 mm x
400.0 mL x 1 L/10³ mL/(0.0821 L·atm/mol·K x 303 K)
n = 0.0152 mol CO2 n = 0.0152 mol CO2 x 44.01 g CO2/ 1 mol CO2
= 0.669 g CO2
![Page 69: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/69.jpg)
⑧Gram molecular mass of NH3 = [N =
1 x 14)] + [H = (3 + 1)] = 14 + 3 = 17 g Molar volume = 22.4 liters Volume of 6.8 g of ammonia at STP
= ?
![Page 70: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/70.jpg)
The ratio between mass and volume is as follows:
17 g : 22.4 liters
6.8 g : x
x = (6.8 x 22.4) / 17 = 8.96 Liters
![Page 71: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/71.jpg)
⑨ 2H2 (g) + O2 (g) → 2H2O (l)
From the equation, 2 volumes of hydrogen react with 1 of oxygen or
2 × 22.4 liters of hydrogen react with 22.4 liters of oxygen.
The volume of hydrogen that will react is 44.8 liters.
![Page 72: Law of Gases](https://reader031.vdocuments.net/reader031/viewer/2022012304/547962bdb4795972098b47bf/html5/thumbnails/72.jpg)
⑩ P = 200 kPa n = 0.250 mol
T = 300K R = 8.314 J K-
1 mol-1
Volume(V) = nRT / P= (0.250 x 8.314 x 300) / 200= 623.55 / 200
V = 3.12 L