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Chapter 6 Gases

6.7d Volume and Moles (Avogadro’s Law)

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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

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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 L2) 1.8 L3) 2.4 L

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Solution

3) 2.4 LSTEP 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

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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)

STP

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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.

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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

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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

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Guide to Using Molar Volume

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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

Learning Check

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A. 1) 5.60 L4.00 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 g8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 L 1 mole He

Solution

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Gases in Equations

The volume or amount of a gas at STP in a chemicalreaction can be calculated from

• STP conditions.

• mole factors from the balanced equation.

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STP and Gas Equations

What volume (L) of O2 gas is needed to completelyreact 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

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What mass of Fe will react with 5.50 L O2 at STP?

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

Learning Check

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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

Solution