IV. Gas Stoichiometry at Non-STP Conditions (p )

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IV. Gas Stoichiometry at Non-STP Conditions (p. 347-350) Ch. 10 & 11 - Gases IV. Gas Stoichiometry at Non-STP Conditions (p. 347-350) C. Johannesson

A. 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. C. Johannesson

B. 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. C. Johannesson

B. Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa =1.017 atm V = ? n = 0.0525 mol T = 25°C = 298 K R = 0.0821 Latm/molK WORK: V = nRT/P V = (0.0525mol)(0.0821Latm/molK)(298K) (1.017 atm) V = 1.26 L CO2 C. Johannesson

B. Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0 L of O2 at 0.9605atm & 21°C? 4 Al + 3 O2  2 Al2O3 15.0 L non-STP ? g GIVEN: P = 0.9605 atm V = 15.0 L n = ? T = 21°C = 294 K R = 0.0821 Latm/molK WORK: PV = nRT (0.9605atm) (15.0 L) = n (0.0821Latm/molK) (294K) n = 0.597 mol O2 Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT  C. Johannesson

B. 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 C. Johannesson