Chapter 13: Gases. Nature of gases Assumptions of Kinetic-Molecular theory are based on four factors: 1)Number of particles present 2)Temperature 3)Pressure.

Chapter 13: Gases

Nature of gases Assumptions of Kinetic-Molecular theory are based on four factors: 1)Number of particles present 2)Temperature 3)Pressure 4)Volume When one variable changes, it affects the other three

Boyle’s Law Boyle’s Law: volume of a given amount of gas held at constant temperature varies inversely with pressure. Increase volume = decrease pressure (less collisions) Decrease volume = increase pressure (more collisions)

Boyle’s Law P 1 V 1 = P 2 V 2 initial final *** You MUST memorize this equation!!!

Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L?

Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? What equation do we use?

Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? What equation do we use? P 1 V 1 = P 2 V 2

Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? Known: Unknown:

Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? Known:V 1 = 4.0 L V 2 = 2.5 L P 1 = 210 kPa Unknown:P 2 = ?

Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? P 1 V 1 = P 2 V 2

Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? (210 kPa) (4.0 L) = (P 2 ) (2.5 L) P 1 V 1 = P 2 V 2

Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? (210 kPa) (4.0 L) = (P 2 ) (2.5 L) 340 kPa = (P 2 ) P 1 V 1 = P 2 V 2

Practice Problem: Boyle’s Law Example: The pressure of a sample of helium in a 1.00-L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00-L container?

Practice Problem: Boyle’s Law Example: The pressure of a sample of helium in a 1.00-L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00-L container? P 1 V 1 = P 2 V 2

Charles’s Law Charles’s Law: volume of a given mass of gas is directly proportional to its kelvin temperature at constant pressure. Increase temperature = Increase volume (faster particles) Decrease temperature = Decrease volume (slower particles)

Charles’s Law V1T1V1T1 V2T2V2T2 = initial final *** You MUST memorize this equation!!!

Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant?

Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? What equation do we use? V1T1V1T1 V2T2V2T2 =

Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? Known: Unknown:

Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? Known: T 1 = 40.0 °C V 1 = 2.32 L T 2 = 75.0 °C Unknown:V 2 = ?

Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? V1T1V1T1 V2T2V2T2 =

Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? V 2 75.0 °C = 2.32 L 40.0 °C

Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? V2V2 = 2.32 L 40.0 °C 75.0 °C x

Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? V2V2 = 2.58 L

Practice Problem: Charles’s Law The celsius temperature of a 3.00-L sample of gas is lowered from 80.0 °C to 30.0 °C. What will be the resulting volume of this gas?

Gay-Lussac’s Law Gay-Lussac’s Law: pressure of a given mass of gas varies directly with kelvin temperature when the volume remains constant. Increase temperature = Increase pressure Decrease temperature = Decrease pressure

Gay-Lussac’s Law P1T1P1T1 P2T2P2T2 = *** You MUST memorize this equation!!!

Using Gay-Lussac’s Law The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank?

Using Gay-Lussac’s Law What equation do we use? The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank?

Using Gay-Lussac’s Law P1T1P1T1 P2T2P2T2 = The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank?

Using Gay-Lussac’s Law Known: Unknown: The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank?

Using Gay-Lussac’s Law Known: T 1 = 22.0 °C P 1 = 3.20 T 2 = 60.0 °C Unknown:P 2 = ? The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank?

Using Gay-Lussac’s Law P1T1P1T1 P2T2P2T2 = The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank?

Using Gay-Lussac’s Law 3.20 atm 22.0 °C P 2 60.0 °C = The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank?

Using Gay-Lussac’s Law 3.20 atm 22.0 °C 60.0 °C X = The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank? P2P2

Using Gay-Lussac’s Law 3.61 atm = The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank? P2P2

Overview of Gas Laws Gas LawTemperaturePressureVolume Boyle’syes Charles’syes Gay-Lussac’syes

14.2 Combined Gas Law We can combine Boyle’s Law, Charles’s Law and Gay- Lussac’s Law into one law (“Combined Gas Law”). States the relationship between pressure, volume, and temperature of a fixed amount of gas. P1V1T1P1V1T1 P2V2T2P2V2T2 =

For the rest of the chapter, we NEED to convert temperature to Kelvin (K) first, BEFORE we use the combined gas law. To convert to Kelvin temperature (K), use the following conversion: T K = 273 + T C Converting to Kelvin (K)

Convert the following temperatures to Kelvin. 1) -25.0 °C 2) 0 °C 3) 23 °C 4) 80.0 °C Converting to Kelvin (K)

Rearrange the Combined Gas Law to isolate the appropriate variable. 1) Solve for P 1 2) Solve for V 1 3) Solve for T 1 4) Solve for T 2 5) Solve for V 2 Solving for a Variable P1V1T1P1V1T1 P2V2T2P2V2T2 =

A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? Using the Combined Gas Law

A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? What equation do we use? Using the Combined Gas Law

A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? What equation do we use? Using the Combined Gas Law P1V1T1P1V1T1 P2V2T2P2V2T2 =

A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? FIRST, what do we do to the temperature values? Using the Combined Gas Law

A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? FIRST, what do we do to the temperature values? CONVERT TO KELVIN! Using the Combined Gas Law

A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? Convert to Kelvin: T 1 = 30.0 °C + 273 = 303 K T 2 = 80.0 °C + 273 = 353 K Using the Combined Gas Law

A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? Known Values: Unknown: Using the Combined Gas Law

A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? Known Values: T 1 = (30.0 °C + 273) 303 K P 1 = 110 kPa V 1 = 2.00 L T 2 = (80.0 °C + 273) 353 K P 2 = 440 kPa Unknown: V 1 = ? Using the Combined Gas Law

A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? T 1 = 303 K P 1 = 110 kPa V 1 = 2.00 L T 2 = 353 K P 2 = 440 kPa V 2 = ? Using the Combined Gas Law P1V1T1P1V1T1 P2V2T2P2V2T2 =

A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? **Plug & Chug Using the Combined Gas Law (110 kPa)(2.00L) 303K (440 kPa)(V 2 ) 353 K =

A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? **Plug & Chug Using the Combined Gas Law (110 kPa)(2.00L) 303K (440 kPa)(V 2 ) 353 K = (V 2 ) =

Example: At 0.00 °C and 1.00 atm pressure, a sample of gas occupies 30.0 mL. If the temperature is increased to 30.0 °C and the entire gas sample is transferred to a 20.0 mL container, what will be the gas pressure inside the container? Practice Problem: Combined Gas Law

Example: At 0.00 °C and 1.00 atm pressure, a sample of gas occupies 30.0 mL. If the temperature is increased to 30.0 °C and the entire gas sample is transferred to a 20.0 mL container, what will be the gas pressure inside the container? What is the first step? What is the second step? Practice Problem: Combined Gas Law

Example: At 0.00 °C and 1.00 atm pressure, a sample of gas occupies 30.0 mL. If the temperature is increased to 30.0 °C and the entire gas sample is transferred to a 20.0 mL container, what will be the gas pressure inside the container? What is the first step? * CONVERT TO KELVIN What is the second step? *Determine Known and Unknown Variables Practice Problem: Combined Gas Law

Example: At 0.00 °C and 1.00 atm pressure, a sample of gas occupies 30.0 mL. If the temperature is increased to 30.0 °C and the entire gas sample is transferred to a 20.0 mL container, what will be the gas pressure inside the container? Practice Problem: Combined Gas Law

The size of large krypton atoms and small helium atoms have no influence on the volume occupied by a fixed number of particles. Avogadro’s Principle: states that equal volumes of gases at the same temperature and pressure contain equal number of particles. Avogadro’s Principle

Molar volume for a gas is the volume that one mole occupies at 0.00 °C (273 K) and 1.00 atm. STP: standard temperature and pressure; 0.00 °C (273 K) and 1.00 atm Avogadro showed that 1 mol gas occupies 22.4 L at STP. Molar Volume

You can use the following conversion factor to find the number of moles, the mass, and even the number of particles in a gas sample. conversion factor: 22.4 L OR 1 mol 1 mol 22.4 L (Oh NO! Conversion Factors!) Molar Volume Conversion Factor

How do you find the number of particles in a sample of gas that has a volume of 3.72 L at STP? Molar Volume

How do you find the number of particles in a sample of gas that has a volume of 3.72 L at STP? Step 1: Calculate the number of moles of gas in 3.72 L. Molar Volume

How do you find the number of particles in a sample of gas that has a volume of 3.72 L at STP? Step 1: Calculate the number of moles of gas in 3.72 L. 3.72 L x Molar Volume

How do you find the number of particles in a sample of gas that has a volume of 3.72 L at STP? Step 1: Calculate the number of moles of gas in 3.72 L. 3.72 L x 1 mol = 0.166 mol 22.4 L Molar Volume

How do you find the number of particles in a sample of gas that has a volume of 3.72 L at STP? Step 2: Convert moles to particles using Avogadro’s number (1 mol = 6.022 x 10 23 particles) Molar Volume

How do you find the number of particles in a sample of gas that has a volume of 3.72 L at STP? Step 2: Convert moles to particles using Avogadro’s number (1 mol = 6.022 x 10 23 particles) 0.166 mol x 6.022x10 23 particles = 9.99x10 22 particles 1 mol Molar Volume

Calculate the volume that 0.861 mol of gas at standard temperature and pressure (STP) will occupy. Avogadro’s Principle

Calculate the volume that 0.861 mol of gas at standard temperature and pressure (STP) will occupy. Hint: Use molar volume conversion factor to calculate the unknown volume. 0.861 mol x Avogadro’s Principle

Calculate the volume that 0.861 mol of gas at standard temperature and pressure (STP) will occupy. Hint: Use molar volume conversion factor to calculate the unknown volume. 0.861 mol x 22.4 L = 19.7 L 1 mol Avogadro’s Principle

Example: How many moles of nitrogen gas will be contained in a 2.00 L flask at STP? Practice Problem:Avogadro’s Principle

Calculate the volume that 2000 g of methane gas (CH 4 ) will occupy at STP. Avogadro’s Principle using Mass

Calculate the volume that 2000 g of methane gas (CH 4 ) will occupy at STP. STEP 1: Convert mass to moles (using molar mass) Avogadro’s Principle using Mass

Calculate the volume that 2000 g of methane gas (CH 4 ) will occupy at STP. STEP 1: Convert mass to moles (using molar mass) 2000 g CH 4 x 1 mol = 125 mol CH 4 16.05 g Avogadro’s Principle using Mass

Calculate the volume that 2000 g of methane gas (CH 4 ) will occupy at STP. STEP 2: Convert moles to volume (since conditions are already at STP, use 22.4 L/mol) 125 mol CH 4 x 22.4 L = 2800 L 1 mol Avogadro’s Principle using Mass

How many grams of carbon dioxide gas are in a 1.0 L balloon at STP? Practice Problems

How many grams of carbon dioxide gas are in a 1.0 L balloon at STP? STEP 1: Convert volume to moles (since conditions are at STP, use 22.4 L/mol) 1.0 L CO 2 x 1 mol = 22.4 L Practice Problems

How many grams of carbon dioxide gas are in a 1.0 L balloon at STP? STEP 2: Convert moles to mass (use molar mass CO 2 ) mol CO 2 x 44.01 g = 1 mol Practice Problems

Combine the laws of Avogadro, Boyle, Charles, and Gay- Lussac into one equation (“The Ideal Gas Law”) to describe the relationship between: -Pressure -Volume -Temperature -Number of moles 14.3 The Ideal Gas Law

The Ideal Gas Law: PV = nRT P = pressure (atm) V = volume (L) n = number of moles (mol) R = gas constant T = temperature (K) The Ideal Gas Law

The Gas Constant, R, depends on the unit of pressure The Ideal Gas Constant, R Units of RNumerical value of R Units of PUnits of VUnits of TUnits of n 0.0821atmLKmol 8.314kPaLKmol 62.4Mm HgLKmol

Ideal Gases: -Particles take up no space and have no intermolecular attractive forces. -Follow all gas laws under all conditions of T & P. Real Gases: -No gases are ideal. -Gas particles have volume (due to size and shape). -Subject to intermolecular forces * Most gases behave like “ideal” gases, except at high pressure and low temps Real vs. Ideal

Calculate the number of moles if gas contained in a 3.0 L vessel at 300 K with a pressure of 1.50 atm. Applying Ideal Gas Law

Calculate the number of moles if gas contained in a 3.0 L vessel at 300 K with a pressure of 1.50 atm. What equation do we use? Applying Ideal Gas Law

Calculate the number of moles if gas contained in a 3.0 L vessel at 300 K with a pressure of 1.50 atm. What equation do we use? Ideal Gas Law! PV = nRT Applying Ideal Gas Law

Calculate the number of moles of gas contained in a 3.0 L vessel at 300 K with a pressure of 1.50 atm. Known: Unknown: Applying Ideal Gas Law

Calculate the number of moles of gas contained in a 3.0 L vessel at 300 K with a pressure of 1.50 atm. Known: V = 3.0 L T = 300 K P = 1.50 atm R = ? Unknown: n = ? Applying Ideal Gas Law

If the pressure exerted by a gas at 25 °C in a volume of 0.044 L is 3.81 atm, how many moles of gas are present? Practice Problem

The Ideal Gas Law can calculate: 1)Moles The Ideal Gas Law can also be used to calculate: 2) Molar mass (if mass of sample is known) 3) Density (if mass of sample is known) Applying the Ideal Gas Law

Ideal Gas Law & Molar Mass

Ideal Gas Law & Density

What is the molar mass of a pure gas that has a density of 1.40 g/L at STP? Ideal Gas Law & Molar Mass

What is the molar mass of a pure gas that has a density of 1.40 g/L at STP? What equation do we use? Hint: given a density, and need to calculate molar mass… Ideal Gas Law & Molar Mass

What is the molar mass of a pure gas that has a density of 1.40 g/L at STP? What equation do we use? Ideal Gas Law & Molar Mass

What is the molar mass of a pure gas that has a density of 1.40 g/L at STP? Known: Unknown: Ideal Gas Law & Molar Mass

How many grams of gas are present in a sample that has a molar mass of 70.0 g/mol and occupies a 2.00-L container at 117 kPa, and 35.1 ° C? Ideal Gas Law & Molar Mass

How many grams of gas are present in a sample that has a molar mass of 70.0 g/mol and occupies a 2.00-L container at 117 kPa, and 35.1 ° C? What equation do we use? Ideal Gas Law & Molar Mass

How many grams of gas are present in a sample that has a molar mass of 70.0 g/mol and occupies a 2.00-L container at 117 kPa, and 35.1 ° C? Known: Unknown: Ideal Gas Law & Molar Mass

How many grams of gas are present in a sample that has a molar mass of 70.0 g/mol and occupies a 2.00-L container at 117 kPa, and 35.1 ° C? Ideal Gas Law & Molar Mass

Gas Laws we learned can be applied to calculate the stoichiometry of reactions w/ gases (either reactants or products) 2C 4 H 10 (g) + 13O 2 (g) → 8CO 2 (g) + 10H 2 O(g) 14.4 Gas Stoichiometry

Coefficients in balanced equation tell us… 2C 4 H 10 (g) + 13O 2 (g) → 8CO 2 (g) + 10H 2 O(g) Gas Stoichiometry

Coefficients in balanced equation tell us… 2C 4 H 10 (g) + 13O 2 (g) → 8CO 2 (g) + 10H 2 O(g) 1)Number of moles 2)Number of representative particles, and … Gas Stoichiometry

Coefficients in balanced equation tell us… 2C 4 H 10 (g) + 13O 2 (g) → 8CO 2 (g) + 10H 2 O(g) 1)Number of moles 2)Number of representative particles 3)Number of Liters! Gas Stoichiometry

How many moles of each gas do we have? 2C 4 H 10 (g) + 13O 2 (g) → 8CO 2 (g) + 10H 2 O(g) ___ mol of C 4 H 10 (g) ___ mol of O 2 (g) ___ mol of CO 2 (g) ___ mol of H 2 O(g) Gas Stoichiometry

How many liters of each gas do we have? 2C 4 H 10 (g) + 13O 2 (g) → 8CO 2 (g) + 10H 2 O(g) ___ L of C 4 H 10 (g) ___ L of O 2 (g) ___ L of CO 2 (g) ___ L of H 2 O(g) Gas Stoichiometry

Moles and Liters in a balanced chemical equation 2C 4 H 10 (g) + 13O 2 (g) → 8CO 2 (g) + 10H 2 O(g) 2 mol 13 mol 8 mol 10 mol 2 L 13 L 8 L 10 L Gas Stoichiometry

Write some conversion factors relating the Liters of gases in the following chemical equation. 2C 4 H 10 (g) + 13O 2 (g) → 8CO 2 (g) + 10H 2 O(g) Gas Stoichiometry with Volume

What volume of methane (CH 4 ) is needed to produce 26 L of water? CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(g) Gas Stoichiometry with Volume

What volume of methane (CH 4 ) is needed to produce 26 L of water? CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(g) Step 1: Write the given!!! Gas Stoichiometry with Volume

What volume of methane (CH 4 ) is needed to produce 26 L of water? CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(g) Step 1: Write the given!!! 26 L H 2 O Gas Stoichiometry with Volume

What volume of methane (CH 4 ) is needed to produce 26 L of water? CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(g) Step 2: Write the conversion factor comparing L of water to L of methane, using volume relationship from coefficients in balanced chemical equation. 26 L H 2 O x Gas Stoichiometry with Volume

What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas (C 3 H 8 )? Assume constant temperature and pressure. Practice Problem: Gas Stoichiometry

What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas (C 3 H 8 )? Assume constant temperature and pressure. C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) Practice Problem: Gas Stoichiometry

What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas (C 3 H 8 )? Assume constant temperature and pressure. C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) Known: Unknown: Practice Problem: Gas Stoichiometry

What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas (C 3 H 8 )? Assume constant temperature and pressure. C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) Known: 4.00 L propane Unknown:? L of oxygen Practice Problem: Gas Stoichiometry

What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas (C 3 H 8 )? Assume constant temperature and pressure. C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) 4.00 L C 3 H 8 x Hint: find Liter to Liter ratio between C 3 H 8 and O 2 Practice Problem: Gas Stoichiometry

Determine the volume of hydrogen gas needed to react completely with 5.00 L of oxygen to form water. Practice Problem: Gas Stoichiometry

We can do stoichiometric calculations involving both gas volumes & masses if we know the following information: 1)Balanced chemical equation 2)At least one mass or volume value for a reactant or product 3)The conditions the gas volume was measured (T & P) *Then use the Ideal Gas Law w/ mole or volume ratios Calculations w/ Volume & Mass

If 5.00 L of nitrogen reacts completely by the following reaction at a constant pressure and temperature of 3.00 atm and 298 K, how many grams of ammonia are produced? N 2 (g) + 3H 2 (g) → 2NH 3 (g) Volume-Mass Problems

If 5.00 L of nitrogen reacts completely by the following reaction at a constant pressure and temperature of 3.00 atm and 298 K, how many grams of ammonia are produced? N 2 (g) + 3H 2 (g) → 2NH 3 (g) Whoa… where do we even begin?!?! Volume-Mass Problems

If 5.00 L of nitrogen reacts completely by the following reaction at a constant pressure and temperature of 3.00 atm and 298 K, how many grams of ammonia are produced? N 2 (g) + 3H 2 (g) → 2NH 3 (g) Analyze the problem…need to convert volume (L) N 2 to mass (g) NH 3 Volume-Mass Problems

If 5.00 L of nitrogen reacts completely by the following reaction at a constant pressure and temperature of 3.00 atm and 298 K, how many grams of ammonia are produced? N 2 (g) + 3H 2 (g) → 2NH 3 (g) Step 1: convert L of N 2 → L of NH 3 (Hint: use coefficients to convert volume to volume) Volume-Mass Problems

If 5.00 L of nitrogen reacts completely by the following reaction at a constant pressure and temperature of 3.00 atm and 298 K, how many grams of ammonia are produced? N 2 (g) + 3H 2 (g) → 2NH 3 (g) Step 1: convert L of N 2 → L of NH 3 (Hint: use coefficients to convert volume to volume) 5.00 L N 2 x Volume-Mass Problems

If 5.00 L of nitrogen reacts completely by the following reaction at a constant pressure and temperature of 3.00 atm and 298 K, how many grams of ammonia are produced? N 2 (g) + 3H 2 (g) → 2NH 3 (g) Step 1: convert L of N 2 → L of NH 3 (Hint: use coefficients to convert volume to volume) 5.00 L N 2 x 2 L NH 3 = 10.0 L NH 3 1 L N 2 Volume-Mass Problems

If 5.00 L of nitrogen reacts completely by the following reaction at a constant pressure and temperature of 3.00 atm and 298 K, how many grams of ammonia are produced? N 2 (g) + 3H 2 (g) → 2NH 3 (g) Step 2: Convert 10.0 L of NH 3 → mol HN 3 (Hint: use PV = nRT, to solve for mol) n = 1.23 mol NH 3 Volume-Mass Problems

If 5.00 L of nitrogen reacts completely by the following reaction at a constant pressure and temperature of 3.00 atm and 298 K, how many grams of ammonia are produced? N 2 (g) + 3H 2 (g) → 2NH 3 (g) Step 3: Convert mol 1.23 mol NH 3 → mass NH 3 (Hint: Use Molar Mass) Volume-Mass Problems

If 5.00 L of nitrogen reacts completely by the following reaction at a constant pressure and temperature of 3.00 atm and 298 K, how many grams of ammonia are produced? N 2 (g) + 3H 2 (g) → 2NH 3 (g) Step 3: Convert mol 1.23 mol NH 3 → mass NH 3 (Hint: Use Molar Mass) 1.23 mol NH 3 x 17.04 g NH 3 = 21.0 g NH 3 1 mol NH 3 Volume-Mass Problems

Use the reaction below to calculate the mass of solid ammonium nitrate that must be used to obtain 0.100 L of dinitrogen oxide gas at STP. NH 4 NO 3 (s) → N2O(g) + 2H 2 O(g) Practice Problem: Volume-Mass

Chapter 13: Gases. Nature of gases Assumptions of Kinetic-Molecular theory are based on four factors: 1)Number of particles present 2)Temperature 3)Pressure.

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Chapter 13: Gases. Nature of gases Assumptions of Kinetic-Molecular theory are based on four factors: 1)Number of particles present 2)Temperature 3)Pressure.

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Presentation on theme: "Chapter 13: Gases. Nature of gases Assumptions of Kinetic-Molecular theory are based on four factors: 1)Number of particles present 2)Temperature 3)Pressure."— Presentation transcript:

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