1.  Bellwork: What is the volume, in liters, of 0.250 mol of oxygen gas at 20.0ºC and 0.974 atm pressure? PV = nRT V = ? n = 0.250 mol T = 20ºC + 273 = 293 K P = 0.947 atm V= nRT P = 0.250 mol (.0821 L·atm/mol·K) (293 K) 0.947 atm = 6.17 L O 2

2. Ch. 11 - Gases IV. Gas Stoichiometry

3. Gas Stoichiometry b Moles  Liters of a Gas: STP - use 22.4 L/mol Non-STP - use ideal gas law b Non- STP Given liters of gas?  start with ideal gas law Looking for liters of gas?  start with stoichiometry conv.

4. 1 mol CaCO 3 100.09g CaCO 3 Gas Stoichiometry Problem b What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? 5.25 g CaCO 3 = 0.0525 mol CO 2 CaCO 3  CaO + CO 2 1 mol CO 2 1 mol CaCO 3 5.25 g? L non-STP Looking for liters: Start with stoich and calculate moles of CO 2. Plug this into the Ideal Gas Law to find liters.

5. WORK: PV = nRTV = nRT/P V=(0.0525mol)(8.315L  kPa/mol  K)(298K) (103 kPa) V = 1.26 L CO 2 Gas Stoichiometry Problem b What volume of CO 2 forms from 5.25 g of CaCO 3 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

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

7. 2 mol Al 2 O 3 3 mol O 2 Gas Stoichiometry Problem b How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? 0.597 mol O 2 = 40.6 g Al 2 O 3 4 Al + 3 O 2  2 Al 2 O 3 101.96 g Al 2 O 3 1 mol Al 2 O 3 15.0L non-STP ? g Use stoich to convert moles of O 2 to grams Al 2 O 3.