Transcript 6_7 Volume and Moles (Avogadro`s Law).

Chapter 6
Gases
6.7
Volume and Moles (Avogadro’s Law)
Copyright © 2009 by Pearson Education, Inc.
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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 of helium gas
occupies a volume of 1.5 L, what
volume will 1.2 moles of helium
occupy at the same temperature
and pressure?
1) 0.94 L
2) 1.8 L
3) 2.4 L
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Solution
3) 2.4 L
STEP 1: Conditions 1
V1 = 1.5 L
n1 = 0.75 mole of He
Conditions 2
V2 = ???
n2 = 1.2 moles of 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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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 mmHg)
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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 write conversion factors.
22.4 L
1 mole
and
1 mole
22.4 L
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Using Molar Volume
What is the volume occupied by 2.75 moles of N2 gas
at STP?
The molar volume is used to convert moles to liters.
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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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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 g of He are present in 8.00 L of gas at STP?
1) 25.6 g
2) 0.357 g
3) 1.43 g
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Solution
A. 1) 5.60 L
4.00 g CH4 x 1 mole CH4
16.0 g CH4
x 22.4 L (STP) = 5.60 L
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
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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.
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Guide to Using Molar Volume
for Reactions
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STP and Gas Equations
What volume (L) of O2 gas at STP is needed to completely
react with 15.0 g of aluminum?
4Al(s) + 3O2(g)
2Al2O3(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 of O2 at STP
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Learning Check
What mass of Fe will react with 5.50 L of O2 at STP?
4Fe(s) + 3O2(g)
2Fe2O3(s)
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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 of Fe
22.4 L
3 moles O2 1 mole Fe
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