1. using The Ideal Gas Law Gas Stoichiometry

2. PV T VnVn PV nT Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.31 L  kPa/mol  K = R

3. Ideal Gas Law UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.31 L  kPa/mol  K PV=nRT

4. GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821 L  atm/mol  K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K P = 3.01 atm Ideal Gas Law b Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW

5. GIVEN: V = ?V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.31 L  kPa/mol  K Ideal Gas Law b Find the volume of 85 g of O 2 at 25°C and 104.5 kPa. = 2.66 mol WORK: 85 g 1 mol = 2.66 mol 32.00 g PV = nRT (104.5)V=(2.66) (8.31) (298) kPa mol L  kPa/mol  K K V = 63.04 L IDEAL GAS LAW

6. Gas Stoichiometry use at non-STP Conditions

7. Gas Stoichiometry b Moles  Liters of a Gas STP  use 22.4 L/mol (Avogadro’s Law) Non-STP  use ideal gas law b Non- STP Problems Given liters of gas?  start with ideal gas law Looking for liters of gas?  start with stoichiometry conversion

8. 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 = 0.052 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 stoichiometry and calculate moles of CO 2. Plug this into the Ideal Gas Law to find liters. x x

9. WORK: PV = nRT (103 kPa)V =(0.052mol)(8.31 L  kPa/mol  K )(298K) V = 1.25 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 = 0.052 mol T = 25°C = 298 K R = 8.31 L  kPa/mol  K

10. WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.31 L  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 = ?n = ? T = 21°C = 294 K R = 8.31 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 

11. 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 = 40.58 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. x x