Presentation on theme: "Volume and Moles (Avogadro’s Law)"— Presentation transcript:
1Volume and Moles (Avogadro’s Law) Chapter Gases6.7dVolume and Moles (Avogadro’s Law)
2Avogadro's Law: Volume and Moles In Avogadro’s Lawthe volume of a gas is directly related to the number of moles (n) of gas.T and P are constant.V1 = V2n n2
3Learning CheckIf 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) L2) 1.8 L3) 2.4 L
4Solution STEP 1 Conditions 1 Conditions 2 V1 = 1.5 L V2 = ??? n1 = mole He n2 = 1.2 moles HeSTEP 2 Solve for unknown V2V2 = V1 x n2n1STEP 3 Substitute values and solve for V2.V2 = 1.5 L x moles He = 2.4 L0.75 mole He
5STPThe volumes of gases can be compared at STP, Standard Temperature and Pressure, when they havethe same temperature.Standard temperature (T)0°C or 273 Kthe same pressure.Standard pressure (P)1 atm (760 mm Hg)
6Molar VolumeAt standard temperature and pressure (STP), 1 mole of a gas occupies a volume of 22.4 L, which is called its molar volume.
7Molar Volume as a Conversion Factor The molar volume at STP can be used to form conversion factors.22.4 L and mole1 mole L
8Using 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 = L1 mole
10Learning Check A. What is the volume at STP of 4.00 g of CH4? 1) L 2) L 3) 44.8 LB. How many grams of He are present in 8.00 L of gas at STP?1) g 2) g 3) 1.43 g
11Solution 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 CH4B. 3) 1.43 g8.00 L x 1 mole He x g He = g He22.4 L mole He
12Gases in Equations The volume or amount of a gas at STP in a chemical reaction can be calculated fromSTP conditions.mole factors from the balanced equation.
13STP and Gas EquationsWhat volume (L) of O2 gas is needed to completelyreact with 15.0 g of aluminum at STP?4 Al(s) O2 (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
14Learning Check 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) What mass of Fe will react with 5.50 L O2 at STP?4 Fe(s) + 3 O2(g) Fe2O3(s)
15Solution4Fe(s) + 3O2(g) Fe2O3(s)? L at STP5.50 L O2 x 1 mole x 4 moles Fe x g Fe = g22.4 L 3 moles O mole Fe