Chapter 6 Gases 7 Volume and Moles (Avogadro’s Law) Avogadro's Law: Volume and Moles


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Chapter 6 Gases

  • 6.7
  • 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.
    • V1 = V2
    • n1 n2

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

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

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)

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.

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

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

Guide to Using Molar Volume

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

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

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.

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 O2 x 22.4 L (STP)
  • 27.0 g Al 4 moles Al 1 mole O2
  • = 9.33 L O2 at STP

Learning Check

  • What mass of Fe will react with 5.50 L O2 at STP?
  • 4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

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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