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Chapter 13: Gases Nature of gases • Assumptions of Kinetic-Molecular theory are based on four factors: 1) 2) 3) 4) Number of particles present Temperature Pressure 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 P1V1 = P2V2 initial *** You MUST memorize this equation!!! final 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? P1V1 = P2V2 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: V1 = 4.0 L V2 = 2.5 L P1 = 210 kPa Unknown: P2 = ? 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? P1V1 = P2V2 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? P1V1 = P2V2 (210 kPa) (4.0 L) = (P2) (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? P1V1 = P2V2 (210 kPa) (4.0 L) = (P2) (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? P1V1 = P2V2 (210 kPa) (4.0 L) = (P2) (2.5 L) 340 kPa = (P2) 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? P1V1 = P2V2 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 V1 V2 = T1 T2 initial *** You MUST memorize this equation!!! final 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? V1 V2 = T1 T2 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: T1 = 40.0 °C V1 = 2.32 L T2 = 75.0 °C Unknown: V2 = ? 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? V1 V2 = T1 T2 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? 2.32 L V2 = 40.0 °C 75.0 °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? 2.32 L 75.0 °C x = 40.0 °C V2 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? 2.58 L = V2 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 P1 P2 = T1 T2 *** 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 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? What equation do we use? 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? P1 P2 = T1 T2 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? Known: Unknown: 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? Known: T1 = 22.0 °C P1 = 3.20 T2 = 60.0 °C Unknown: P2 = ? 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? P1 P2 = T1 T2 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? 3.20 atm P2 = 22.0 °C 60.0 °C 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? 60.0 °C X 3.20 atm = P2 22.0 °C 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? 3.61 atm = P2 Overview of Gas Laws Gas Law Temperature Boyle’s Charles’s yes Gay-Lussac’s yes Pressure Volume yes yes yes yes 14.2 Combined Gas Law • We can combine Boyle’s Law, Charles’s Law and GayLussac’s Law into one law (“Combined Gas Law”). • States the relationship between pressure, volume, and temperature of a fixed amount of gas. P1V1 P2V2 = T1 T2 Converting to Kelvin (K) • 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: TK = 273 + TC Converting to Kelvin (K) Convert the following temperatures to Kelvin. 1) -25.0 °C 2) 0 °C 3) 23 °C 4) 80.0 °C Solving for a Variable Rearrange the Combined Gas Law to isolate the appropriate variable. 1) Solve for P1 2) Solve for V1 3) Solve for T1 4) Solve for T2 5) Solve for V2 P1V1 P2V2 = T1 T2 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? 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? P1V1 P2V2 = T1 T2 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? 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: T1 = 30.0 °C + 273 = 303 K T2 = 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: T1 = (30.0 °C + 273) 303 K P1 = 110 kPa V1 = 2.00 L T2 = (80.0 °C + 273) 353 K P2 = 440 kPa Unknown: V1 = ? 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? T1 = 303 K P1 = 110 kPa V1 = 2.00 L T2 = 353 K P2 = 440 kPa V2 = ? P1V1 P2V2 = T1 T2 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? **Plug & Chug (110 kPa)(2.00L) 303K = (440 kPa)(V2) 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? **Plug & Chug (110 kPa)(2.00L) 303K = (V2) = (440 kPa)(V2) 353 K 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 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? Avogadro’s Principle • 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. Molar Volume • 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 Conversion Factor • 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 1 mol (Oh NO! Conversion Factors!) OR 1 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? 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 1023 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 1023 particles) 0.166 mol x 6.022x1023 particles = 9.99x1022 particles 1 mol Avogadro’s Principle 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 Practice Problem:Avogadro’s Principle Example: How many moles of nitrogen gas will be contained in a 2.00 L flask at STP? Avogadro’s Principle using Mass Calculate the volume that 2000 g of methane gas (CH4) will occupy at STP. Avogadro’s Principle using Mass Calculate the volume that 2000 g of methane gas (CH4) 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 (CH4) will occupy at STP. STEP 1: Convert mass to moles (using molar mass) 2000 g CH4 x 1 mol = 125 mol CH4 16.05 g Avogadro’s Principle using Mass Calculate the volume that 2000 g of methane gas (CH4) will occupy at STP. STEP 2: Convert moles to volume (since conditions are already at STP, use 22.4 L/mol) 125 mol CH4 x 22.4 L 1 mol = 2800 L Practice Problems 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 CO2 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 CO2) mol CO2 x 44.01 g 1 mol = 14.3 The Ideal Gas Law Combine the laws of Avogadro, Boyle, Charles, and GayLussac into one equation (“The Ideal Gas Law”) to describe the relationship between: - Pressure Volume Temperature Number of moles 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 Constant, R The Gas Constant, R, depends on the unit of pressure Units of R Numerical value of R Units of P Units of V Units of T Units of n 𝐿 · 𝒂𝒕𝒎 𝑚𝑜𝑙 · 𝐾 0.0821 atm L K mol 𝐿 · 𝒌𝑷𝒂 𝑚𝑜𝑙 · 𝐾 8.314 kPa L K mol 𝐿 · 𝒎𝒎 𝑯𝒈 𝑚𝑜𝑙 · 𝐾 62.4 Mm Hg L K mol Real vs. Ideal 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 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. 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 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 𝐿 · 𝑎𝑡𝑚 R = 0.0821 𝑚𝑜𝑙 ·𝐾 Unknown: n = ? 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. PV = nRT (1.50 atm)(3.0 L) = (1.5 atm)(3.0 L) (0.0821 𝐿 ·𝑎𝑡𝑚 )(300 𝑚𝑜𝑙 ·𝐾 𝐿 ·𝑎𝑡𝑚 n(0.0821 )(300 K) 𝑚𝑜𝑙 ·𝐾 =n K) n = 0.18 mol Practice Problem 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? Applying the Ideal Gas Law 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) Ideal Gas Law & Molar Mass Relationship between moles and molar mass: moles (n) = PV = nRT 𝑚𝑎𝑠𝑠 (𝑚) 𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 (𝑀) changes to rearrange to solve for M, ; 𝑚 (n = ) 𝑀 𝑚𝑅𝑇 PV = 𝑀 𝑚𝑅𝑇 M = 𝑃𝑉 Ideal Gas Law & Density Relationship between moles and density: 𝑚𝑎𝑠𝑠 (𝑚) Density (D) = ; 𝑉𝑜𝑙𝑢𝑚𝑒 (𝐿) 𝑚𝑅𝑇 M= 𝑃𝑉 (D = 𝑚 𝑉 changes to 𝐷𝑅𝑇 M = 𝑃 Rearrange to solve for D, 𝑀𝑃 D = 𝑅𝑇 ) 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? 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? M = 𝐷𝑅𝑇 𝑃 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? M = Known: Unknown: 𝐷𝑅𝑇 𝑃 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? M = Known: D: 1.40 g/L T: ? P: ? 𝐿 ·𝑎𝑡𝑚 R: 0.0821 𝑚𝑜𝑙 ·𝐾 Unknown: M: ? 𝐷𝑅𝑇 𝑃 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? M = Known: D: 1.40 g/L T: 273 K P: 1 atm 𝐿 ·𝑎𝑡𝑚 R: 0.0821 𝑚𝑜𝑙 ·𝐾 Unknown: M: ? 𝐷𝑅𝑇 𝑃 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? M = 𝐷𝑅𝑇 𝑃 𝑔 M = 𝐿 ·𝑎𝑡𝑚 (1.40 𝐿 )(0.0821𝑚𝑜𝑙 ·𝐾)(273 𝐾) (1𝑎𝑡𝑚) M = 31.4 g/mol 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? What equation do we use? PV = 𝑚𝑅𝑇 𝑀 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? PV = Known: Unknown: 𝑚𝑅𝑇 𝑀 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? PV = Known: 𝑚𝑅𝑇 𝑀 M: 70.0 g/mol V: 2.00 L P: 117 kPa T: (35.1 °C + 273) = 308.1 K 𝐿 ·𝑘𝑃𝑎 R: 8.314 𝑚𝑜𝑙 ·𝐾 Unknown: m: ? 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? m= 𝑃𝑉𝑀 𝑅𝑇 14.4 Gas Stoichiometry Gas Laws we learned can be applied to calculate the stoichiometry of reactions w/ gases (either reactants or products) 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) Gas Stoichiometry Coefficients in balanced equation tell us… 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) Gas Stoichiometry Coefficients in balanced equation tell us… 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) 1) Number of moles 2) Number of representative particles, and … Gas Stoichiometry Coefficients in balanced equation tell us… 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(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? 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) ___ mol of C4H10(g) ___ mol of O2(g) ___ mol of CO2(g) ___ mol of H2O(g) Gas Stoichiometry How many liters of each gas do we have? 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) ___ L of C4H10(g) ___ L of O2(g) ___ L of CO2(g) ___ L of H2O(g) Gas Stoichiometry Moles and Liters in a balanced chemical equation 2C4H10(g) 2 mol 2L + 13O2(g) 13 mol 13 L → 8CO2(g) 8 mol 8L + 10H2O(g) 10 mol 10 L Gas Stoichiometry with Volume Write some conversion factors relating the Liters of gases in the following chemical equation. 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) Gas Stoichiometry with Volume What volume of methane (CH4) is needed to produce 26 L of water? CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) Gas Stoichiometry with Volume What volume of methane (CH4) is needed to produce 26 L of water? CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) Step 1: Write the given!!! Gas Stoichiometry with Volume What volume of methane (CH4) is needed to produce 26 L of water? CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) Step 1: Write the given!!! 26 L H2O Gas Stoichiometry with Volume What volume of methane (CH4) is needed to produce 26 L of water? CH4(g) + 2O2(g) → CO2(g) + 2H2O(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 H2O x Practice Problem: Gas Stoichiometry What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas (C3H8)? 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 (C3H8)? Assume constant temperature and pressure. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Practice Problem: Gas Stoichiometry What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas (C3H8)? Assume constant temperature and pressure. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(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 (C3H8)? Assume constant temperature and pressure. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(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 (C3H8)? Assume constant temperature and pressure. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) 4.00 L C3H8 x Hint: find Liter to Liter ratio between C3H8 and O2 Practice Problem: Gas Stoichiometry Determine the volume of hydrogen gas needed to react completely with 5.00 L of oxygen to form water. Calculations w/ Volume & Mass 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 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? N2(g) + 3H2(g) → 2NH3(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? N2(g) + 3H2(g) → 2NH3(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? N2(g) + 3H2(g) → 2NH3(g) Analyze the problem…need to convert volume (L) N2 to mass (g) NH3 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? N2(g) + 3H2(g) → 2NH3(g) Analyze the problem…need to convert volume (L) N2 to mass (g) NH3 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? N2(g) + 3H2(g) → 2NH3(g) Step 1: convert L of N2 → L of NH3 (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? N2(g) + 3H2(g) → 2NH3(g) Step 1: convert L of N2 → L of NH3 (Hint: use coefficients to convert volume to volume) 5.00 L N2 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? N2(g) + 3H2(g) → 2NH3(g) Step 1: convert L of N2 → L of NH3 (Hint: use coefficients to convert volume to volume) 5.00 L N2 x 2 L NH3 = 10.0 L NH3 1 L N2 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? N2(g) + 3H2(g) → 2NH3(g) Step 2: Convert 10.0 L of NH3 → mol HN3 (Hint: use PV = nRT, to solve for mol) 𝑃𝑉 n= 𝑅𝑇 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? N2(g) + 3H2(g) → 2NH3(g) Step 2: Convert 10.0 L of NH3 → mol HN3 (Hint: use PV = nRT, to solve for mol) V: 10.0 L P: 3.00 atm T: 298 K 𝐿·𝑎𝑡𝑚 R: 0.0821 𝑚𝑜𝑙·𝐾 n= 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? N2(g) + 3H2(g) → 2NH3(g) Step 2: Convert 10.0 L of NH3 → mol HN3 (Hint: use PV = nRT, to solve for mol) n = 1.23 mol NH3 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? N2(g) + 3H2(g) → 2NH3(g) Step 3: Convert mol 1.23 mol NH3 → mass NH3 (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? N2(g) + 3H2(g) → 2NH3(g) Step 3: Convert mol 1.23 mol NH3 → mass NH3 (Hint: Use Molar Mass) 1.23 mol NH3 x 17.04 g NH3 = 21.0 g NH3 1 mol NH3 Practice Problem: Volume-Mass 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. NH4NO3(s) → N2O(g) + 2H2O(g)