7.7 Volume and Moles (Avogadro’s Law) Chapter 7 Gases 7.7 Volume and Moles (Avogadro’s Law)

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 n1 n2

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

Solution 3) 2.4 L STEP 1 V1 = 1.5 L V2 = ? V increases Conditions 1 Conditions 2 Know Predict V1 = 1.5 L V2 = ? V increases n1 = 0.75 mole n2 = 1.2 moles n increases 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

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)

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

Molar Volume as a Conversion Factor The molar volume at STP has about the same volume as 3 basketballs can be used to form 2 conversion factors: 22.4 L and 1 mole 1 mole 22.4 L

Guide to Using Molar Volume

Using Molar Volume What is the volume occupied by 2.75 moles of N2 gas at STP? STEP 1 Given: 2.75 moles of N2 Need: Liters of N2 STEP 2 Write a plan: Use the molar volume to convert moles to liters.

Using Molar Volume (continued) STEP 3 Write equalities and conversion factors: 1 mole of gas = 22.4 L 1 mole gas and 22.4 L 22.4 L 1 mole gas STEP 4 Substitute data and solve: 2.75 moles N2 x 22.4 L = 61.6 L of N2 1 mole N2

Learning Check What is the volume at STP of 4.00 g of CH4? 1) 5.60 L 2) 11.2 L 3) 44.8 L

Solution 1) 5.6 L STEP 1 Given: 4.00 g of CH4 Need: liters of CH4 at STP STEP 2 Write a plan: g of CH4 moles of CH4 L of CH4 (at STP) STEP 3 Write equalities and conversion factors: 1 mole of CH4 1 mole of gas = 22.4 L 1 mole CH4 and 16.0 g CH4 1 mole gas and 22.4 L 16.0 g CH4 1 mole CH4 22.4 L 1 mole gas STEP 4 Substitute data and solve: 4.00 g CH4 x 1 mole CH4 x 22.4 L = 5.6 L of CH4 (STP) 16.0 g CH4 1 mole CH4

Learning Check How many grams of He are present in 8.00 L of He at STP? 1) 25.6 g 2) 0.357 g 3) 1.43 g

Solution 3) 1.43 g of He STEP 1 Given: 8.00 L of He Need: grams of He at STP STEP 2 Write a plan: L of He moles of He g of He (at STP) STEP 3 Write equalities and conversion factors: 1 mole of gas = 22.4 L 1 mole of He = 4.00 g of He 1 mole gas and 22.4 L 1 mole He and 4.00 g He 22.4 L 1 mole gas 4.00 g He 1 mole CH4 STEP 4 Substitute data and solve: 8.00 L He x 1 mole He x 4.00 g He = 1.43 g of He 22.4 L He 1 mole He

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

Guide to Problem Solving Reactions Involving Gases

Example of Gases in Equations What volume (L) of O2 gas is needed to completely react with 15.0 g of aluminum at STP? 4Al(s) + 3O2 (g) 2Al2O3(s)

Solution for Example of Gases in Equations What volume (L) of O2 gas is needed to completely react with 15.0 g of aluminum at STP? 4Al(s) + 3O2 (g) 2Al2O3(s) STEP 1 Find moles of Al using molar mass. STEP 2 Determine moles of O2 using mole–mole factor. STEP 3 Convert moles of O2 to L using molar volume. (STEP 1) (STEP 2) (STEP 3) 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)

Learning Check What mass of Fe will react with 5.50 L of O2 at STP? 4Fe(s) + 3O2(g) 2Fe2O3(s) 1) 13.7 g of Fe 2) 18.3 g of Fe 3) 419 g of Fe

Solution STEP 1 Find moles of O2 using molar volume. STEP 2 Determine moles of Fe using mole–mole factor. STEP 3 Convert moles of Fe to grams using molar mass. (STEP 1) (STEP 2) (STEP 3) 5.50 L O2 x 1 mole O2 x 4 moles Fe x 55.9 g Fe 22.4 L O2 3 moles O2 1 mole Fe = 18.3 g of Fe