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Bellwork:

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 O 2. =. nRT P. 0.250 mol (.0821 L·atm/mol·K) (293 K) 0.947 atm. V. =. =. Ch. 11 - Gases.

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Bellwork:

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  1. 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 = =

  2. Ch. 11 - Gases IV. Gas Stoichiometry

  3. Gas Stoichiometry • Moles  Liters of a Gas: • 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.

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

  5. 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=8.315L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

  6. 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=8.315L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 

  7. 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

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