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Gases and the Mole. Gay-Lussac formulated the law of combining volumes. Law of Combining Volumes. Gases at the same temperature and pressure react with one another in volume ratios of small whole numbers. Law of Combining Volumes. Example 1:. H 2 + Cl 2. 2HCl. 1 L + 1 L. 2 L.
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Law of Combining Volumes Gases at the same temperature and pressure react with one another in volume ratios of small whole numbers.
Law of Combining Volumes Example 1: H2 + Cl2 2HCl 1 L + 1 L 2 L
Law of Combining Volumes Example 2: 2H2 + O2 2H2O 2 L + 1 L 2 L
Avogadro’s law The volume of a gas, maintained at a constant temperature and pressure, is directly proportional to the number of moles of the gas.
Molar Volume the volume that a mole of gas occupies at standard temperature and pressure
Molar Volume 1 mole of ANY gas at STP occupies 22.4 L.
Sample Problem 1 What volume would 7 moles of carbon dioxide occupy at STP? 7 mol CO2 22.4 L = 157 L CO2 1 mol
Question Change 228 L of O2 at STP to moles. • 470 moles • 426 moles • 10,035 moles • 10 moles
228 L 1 mol = 10.2 mol 22.4 L
Sample Problem 2 What volume will 4.00 mol of ammonia occupy at STP? 4.00 mol NH3 22.4 L = 89.6 L NH3 at STP 1 mol
Sample Problem 3 What volume will 2.50 mol of hydrogen gas occupy at 300. K and 400. torr? 2.50 mol H2 22.4 L = 56.0 L H2 at STP 1 mol
Sample Problem 3 = P1V1 P2V2 T1 T2 The combined gas law is used to adjust this volume to nonstandard conditions.
Sample Problem 3 (760. torr) (56.0 L) (400. torr) (V2) = 273 K 300. K (760. torr) (56.0 L) (300. K) = V2 (273 K) (400. torr) 117 L = V2
Sample Problem 4 A sample of oxygen gas occupies 1.00 L when its temperature is 190. K and its pressure is 129 kPa. How many moles of oxygen are present?
Sample Problem 4 = P1V1 P2V2 T1 T2 The given solution must first be adjusted to standard conditions.
Sample Problem 4 (129 kPa) (1.00 L) (101.3 kPa) (V2) = 190. K 273 K (129 kPa) (1.00 L) (273 K) = V2 (190. K) (101.3 kPa) V2 = 1.83 L at STP
Sample Problem 4 Once the gas’s volume at STP has been calculated, the number of moles can be found. 1.83 L 1 mol O2 = 0.0817 mol O2 22.4 L
1 mol 1 mol Every gas has the same number of particles in one mole; but, the mass of each gas is different.
Gas Density Recall that density is mass/volume. For a gas, its density would have units of grams/liter.
Gas Density We can expand this definition to include moles: grams/mole L/mole D =
Gas Density However, what is the number of liters/mole for all gasses at STP? 22.4 Therefore, the denominator is always 22.4 (at STP).
Sample Problem 5 grams/mole L/mole D = What is the density of the gas Cl2 at STP? 70.9 grams/mole 22.4 L/mole D = = 3.17g/L
Sample Problem 6 molar mass molar volume D of H2 = What is the density of hydrogen gas? 2.016 g/mol 22.4 L/mol = 0.0900 g/L
Sample Problem 7 molar mass density molar volume of O2 = If the density of oxygen gas at STP is 1.429 g/L, find the molar volume of the gas. 32.00 g/mol 1.429 g/L = 22.39 L/mol
Sample Problem 8 molarmass molarvolume = density × One mole of an unknown gas has a density of 2.1444 g/L at STP. What is its molar mass? = 2.144 g/L × 22.4 L/mol = 48.0 g/mol
Sample Problem 8 One mole of an unknown gas has a density of 2.1444 g/L at STP. What is its molar mass? Further tests could prove this gas to be ozone, which has a molar mass of 48.00 g/L.
Stoichiometry The “Math” of Chemistry
“Skeleton Equation” A + B C + D Start with “X” grams of A. How many grams of B? X g A 1 mol A # mol B # g B # mol A 1 mol B # g A
Question How do you change from moles of one substance to moles of another substance? • Periodic table • Coefficient ratio • Combined gas law • Number of particles
Question How do you change from grams to moles? • Periodic table • Coefficient ratio • Combined gas law • Number of particles
1 10 2 9 3 8 4 7 5 6
Question How do you change from one set of conditions to another? • Combined gas law • Coefficient ratio • Number of particles • Use 22.4
Question How do you change from volume of gas to moles of gas? • Combined gas law • Coefficient ratio • Number of particles • Use 22.4
Sample Problem 9 When 2.00 mol of calcium react with water to form calcium hydroxide and hydrogen gas, what volume of hydrogen gas will be produced at STP?
Sample Problem 9 Solution: First, write out the reaction and balance it. Ca + 2 H2O Ca(OH)2 + H2 Moles of calcium can be used to find moles of hydrogen, which can then be converted to the volume of hydrogen.
Sample Problem 9 2.00 mol Ca 1 mol H2 22.4 L 1 mol Ca 1 mol H2 = 44.8 L H2 at STP
Sample Problem 10 How many grams of water will be produced if 0.500 L of oxygen gas at STP is burned with hydrogen? Solution: The balanced equation for the reaction is 2H2 + O2 2H2O.
Sample Problem 10 The volume of oxygen must be converted to the moles of oxygen. The moles of oxygen can be used to find moles of water, which can then be converted to the mass of water.
Sample Problem 10 0.500 L O2 1 mol O2 2 mol H2O 22.4 L O2 1 mol O2 18.016 g H2O = 0.804 g H2O 1 mol H2O
Sample Problem 11 If 15.9 g of Xenon gas reacts with excess fluorine, how many liters of xenon hexafluoride will be produced at 500°C? Xe + 3 F2 XeF6
Plan: gXe mXe mXeF6 lXeF6 (at STP) lXeF6 (at STP) lXeF6 (at 500°C)
15.9 g Xe 1 mol Xe 1 mol XeF6 131.3 g Xe 1 mol Xe 22.4 L XeF6 = 2.71 L XeF6 1 mol XeF6
2.71 L XeF6 V2 = 273 K 773 K V2 = 7.67 L XeF6 at 500 °C
Sample Problem 12 For 2 H2 + O2 2 H2O, if you start with 45 L of H2 at 300 K and 700 mm Hg, how many g of O2 will be needed? Plan: Change H2 to STP(combined gas law) Change H2 to g O2(unit analysis)
(700 mmHg)(45 L) (300 K) (760 mmHg)V2 (273 K) = (700)(45 L)(273) = (300)(760)V2 V2 = 37.7 L H2
37.7L H2 1mol 22.4L 1 molO2 2 molH2 32 gO2 1 mol O2 = 26.9 g O2
Sample Problem 13 When 17 grams of calcium react with water, Ca + 2 H2O(l) Ca(OH)2 + H2. What volume of H2 will be produced at 300 K and a pressure of 560 mm Hg?
Plan: Use “skeleton equation” to change to L of H2 at STP. Use combined gas law to change to L of H2 not at STP.