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

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