1. C. Johannesson Ch. 10 & 11 - Gases Gas Stoichiometry at Non- STP Conditions

2. C. Johannesson A. 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.

3. C. Johannesson 1 mol CaCO 3 100.09g CaCO 3 B. 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 = 1.26 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.

4. C. Johannesson WORK: PV = nRT V = B. 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 = ? atm V = ? n = 1.26 mol T = 25°C = 298 K R =.0821 L atm/mol K

5. C. Johannesson WORK: PV = nRT ( atm) (15.0 L) = n (.0821) (294K) n = 0.597 mol O 2 B. 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 =.0821 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 

6. C. Johannesson 2 mol Al 2 O 3 3 mol O 2 B. 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.