Bellwork: What is the volume, in liters, of 0.250 mol of oxygen gas at 20.0ºC and 0.974 atm pressure? V = ? n = 0.250 mol T = 20ºC + 273 = 293 K P = 0.947 atm PV = nRT 6.17 L O2 nRT P 0.250 mol (.0821 L·atm/mol·K) (293 K) 0.947 atm = V = =

Gas Stoichiometry at Non-STP Conditions

OBJECTIVES 4.2.11 Use a chemical equation to specify volume ratios for gaseous reactants or products, or both. 4.2.12 Use volume ratios and the gas laws to calculate volumes, masses, or molar amounts of gaseous reactants or products.

Gas Stoichiometry Moles  Liters of a Gas: Non-STP STP - use 22.4 L/mol Non-STP - use ideal gas law Non-STP Given liters of gas? start with ideal gas law Looking for liters of gas? start with stoichiometry conv. If given L of gas and are looking for L of a different gas, just use the mole ratio!

Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? CaCO3  CaO + CO2 5.25 g ? L non-STP Looking for liters: Start with stoich and calculate moles of CO2. 5.25 g CaCO3 1 mol CaCO3 100.09g 1 mol CO2 CaCO3 = 0.0525 mol CO2 Plug this into the Ideal Gas Law to find liters.

Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 0.0525 mol T = 25°C = 298 K R = 8.315 LkPa/molK WORK: PV = nRT V = nRT/P V=(0.0525mol)(8.315LkPa/molK)(298K) (103 kPa) V = 1.26 L CO2

Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al + 3 O2  2 Al2O3 15.0 L non-STP ? g GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = 8.315 LkPa/molK WORK: PV = nRT n = PV/RT n= (97.3 kPa) (15.0 L) (8.315LkPa/molK) (294K) n = 0.597 mol O2 Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT 

Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al + 3 O2  2 Al2O3 15.0L non-STP ? g Use stoich to convert moles of O2 to grams Al2O3. 0.597 mol O2 2 mol Al2O3 3 mol O2 101.96 g Al2O3 1 mol Al2O3 = 40.6 g Al2O3

CaCO3(s) CaO(s) + CO2(g) ? g 5.00 L How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide gas at STP? V = 5.00 L T = 0ºC + 273 = 273 K P = 1 atm n = ? At STP 1 mole = 22.4 L 5.00 L n 1 mol 22.4 L .223 mol CO2 = x = 1 mol CaCO3 1 mol CO2 100.09 g CaCO3 1 mol CaCO3 22.3 g CaCO3 .223 mol CO3 = x x

WO3(s) + 3H2(g) W(s) + 3H2O(g) ? L How many liters of hydrogen gas at 35ºC and 0.980 atm are needed to react completely with 875 g of tungsten oxide? 1 mol WO3 231.84 g WO3 3 mol H2 1 mol WO3 11.3 mol H2 = 875 g WO3 x x n = 11.3 mol P = 0.980 atm T = 35ºC + 273 = 308 K V = ? PV = nRT nRT P 11.3 mol (.0821 L·atm/mol·K) (308 K) (0.980 atm) V = 292 L H2 = =