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Gases Volume and Moles (Avogadro’s Law)

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Presentation on theme: "Gases Volume and Moles (Avogadro’s Law)"— Presentation transcript:

1 Gases Volume and Moles (Avogadro’s Law)
Copyright © by Pearson Education, Inc. Publishing as Benjamin Cummings Edited by bbg

2 Avogadro's Law: Volume and Moles
Avogadro’s Law states that The volume of a gas is directly related to the number of moles (n) of gas. T and P are constant. V1 = V2 n n2

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) L 2) 1.8 L 3) 2.4 L

4 Solution 2.4 L STEP 1 Conditions 1 Conditions 2 V1 = 1.5 L V2 = ???
n1 = 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 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 The molar volume of a gas
Is measured at STP (standard temperature and pressure). Is 22.4 L for 1 mole of any gas. .

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

9 Guide to Using Molar Volume
Copyright © by Pearson Education, Inc. Publishing as Benjamin Cummings

10 Learning Check A. What is the volume at STP of 4.00 g of CH4?
1) L 2) L 3) 44.8 L B. How many grams of He are present in 8.00 L of Hes at STP? 1) g 2) 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 CH mole CH4 B. 3) 1.43 g 8.00 L x 1 mole He x g He = g He 22.4 L mole He

12 Gases in Chemical Reactions
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? 4Al(s) + 3O2 (g) Al2O3(s) Plan: g Al mole Al mole O L O2 (STP) 15.0 g Al x 1 mole Al x 3 moles O2 x L (STP) 27.0 g Al moles Al mole O2 = L O2 at STP

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

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

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