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1 Chapter 6 Gases 6.7d Volume and Moles (Avogadro’s Law)

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Presentation on theme: "1 Chapter 6 Gases 6.7d Volume and Moles (Avogadro’s Law)"— Presentation transcript:

1 1 Chapter 6 Gases 6.7d Volume and Moles (Avogadro’s Law)

2 2 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. V 1 = V 2 n 1 n 2

3 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) 0.94 L 2)1.8 L 3) 2.4 L

4 4 Solution 3) 2.4 L STEP 1 Conditions 1Conditions 2 V 1 = 1.5 LV 2 = ??? n 1 = 0.75 mole Hen 2 = 1.2 moles He STEP 2 Solve for unknown V 2 V 2 = V 1 x n 2 n 1 STEP 3 Substitute values and solve for V 2. V 2 = 1.5 L x 1.2 moles He = 2.4 L 0.75 mole He

5 5 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

6 6 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.

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

8 8 Using Molar Volume What is the volume occupied by 2.75 moles N 2 gas at STP? The molar volume is used to convert moles to liters moles N 2 x 22.4 L = 61.6 L 1 mole

9 9 Guide to Using Molar Volume

10 10 A. What is the volume at STP of 4.00 g of CH 4 ? 1) 5.60 L2) 11.2 L3) 44.8 L B. How many grams of He are present in 8.00 L of gas at STP? 1) 25.6 g2) g3) 1.43 g Learning Check

11 11 A. 1) 5.60 L 4.00 g CH 4 x 1 mole CH 4 x 22.4 L (STP) = 5.60 L 16.0 g CH 4 1 mole CH 4 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 Solution

12 12 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.

13 13 STP and Gas Equations What volume (L) of O 2 gas is needed to completely react with 15.0 g of aluminum at STP? 4 Al(s) + 3 O 2 (g) 2 Al 2 O 3 (s) Plan: g Al mole Al mole O2 L O2 (STP) 15.0 g Al x 1 mole Al x 3 moles O 2 x 22.4 L (STP) 27.0 g Al 4 moles Al 1 mole O 2 = 9.33 L O 2 at STP

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

15 15 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) ? 5.50 L at STP 5.50 L O 2 x 1 mole x 4 moles Fe x 55.9 g Fe = 18.3 g 22.4 L 3 moles O 2 1 mole Fe Solution


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