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Chapter 6 Gases

Chapter 6 Gases. 6.7d Volume and Moles (Avogadro’s Law). Avogadro's Law: Volume and Moles. In Avogadro’s Law the volume of a gas is directly related to the number of moles ( n ) of gas. T and P are constant. V 1 = V 2 n 1 n 2. Learning Check.

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Chapter 6 Gases

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  1. Chapter 6 Gases 6.7d Volume and Moles (Avogadro’s Law)

  2. Avogadro's Law: Volume and Moles In Avogadro’s Law • the volume of a gas is directly related to the number of moles (n) of gas. • T and P are constant. V1 = V2 n1n2

  3. Learning Check If 0.75 mole helium gas occupies a volume of 1.5 L, what volume will 1.2 moles helium occupy at the same temperature and pressure? 1) 0.94 L 2) 1.8 L 3) 2.4 L

  4. Solution 3) 2.4 L STEP 1 Conditions 1 Conditions 2 V1 = 1.5 L V2 = ??? n1 = 0.75 mole He n2 = 1.2 moles He STEP 2 Solve for unknown V2 V2 = V1 x n2 n1 STEP 3 Substitute values and solve for V2. V2 = 1.5 L x 1.2 moles He = 2.4 L 0.75 mole He

  5. STP The volumes of gases can be compared at STP, Standard Temperature and Pressure, when they have • the same temperature. Standard temperature (T) 0°C or 273 K • the same pressure. Standard pressure (P) 1 atm (760 mm Hg)

  6. Molar Volume At standard temperature and pressure (STP), 1 mole of a gas occupies a volume of 22.4 L, which is called its molar volume.

  7. Molar Volume as a Conversion Factor The molar volume at STP can be used to form conversion factors. 22.4 L and 1 mole 1 mole 22.4 L

  8. Using Molar Volume What is the volume occupied by 2.75 moles N2 gas at STP? The molar volume is used to convert moles to liters. 2.75 moles N2 x 22.4 L = 61.6 L 1 mole

  9. Guide to Using Molar Volume

  10. Learning Check A. What is the volume at STP of 4.00 g of CH4? 1) 5.60 L 2) 11.2 L 3) 44.8 L B. How many grams of He are present in 8.00 L of gas at STP? 1) 25.6 g 2) 0.357 g 3) 1.43 g

  11. Solution A. 1) 5.60 L 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L 16.0 g CH4 1 mole CH4 B. 3) 1.43 g 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 L 1 mole He

  12. Gases in Equations The volume or amount of a gas at STP in a chemical reaction can be calculated from • STP conditions. • mole factors from the balanced equation.

  13. STP and Gas Equations What volume (L) of O2 gas is needed to completely react with 15.0 g of aluminum at STP? 4 Al(s) + 3 O2 (g) 2 Al2O3(s) Plan: g Al mole Al mole O2 L O2 (STP) 15.0 g Al x 1 mole Al x 3 moles O2x 22.4 L (STP) 27.0 g Al 4 moles Al 1 mole O2 = 9.33 L O2 at STP

  14. Learning Check What mass of Fe will react with 5.50 L O2 atSTP? 4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

  15. Solution 4Fe(s) + 3O2(g) 2Fe2O3(s) ? 5.50 L at STP 5.50 L O2 x 1 mole x 4 moles Fe x 55.9 g Fe = 18.3 g 22.4 L 3 moles O2 1 mole Fe

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