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**Gases Volume and Moles (Avogadro’s Law)**

Copyright © by Pearson Education, Inc. Publishing as Benjamin Cummings Edited by bbg

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**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

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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) L 2) 1.8 L 3) 2.4 L

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**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

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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 mm Hg)

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**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. .

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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 mole 1 mole 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 L = L 1 mole

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**Guide to Using Molar Volume**

Copyright © by Pearson Education, Inc. Publishing as Benjamin Cummings

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**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

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**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

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**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.

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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

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**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)

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**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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