Gas Stoichiometry at Non-STP Conditions

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Presentation transcript:

Gas Stoichiometry at Non-STP Conditions Gases Gas Stoichiometry at Non-STP Conditions

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.

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 = .0.052 mol CO2 Plug this into the Ideal Gas Law to find liters.

B. Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 0.052 mol T = 25°C = 298 K R = 8.315 dm3kPa/molK WORK: PV = nRT (103 kPa)V =(.052mol)(8.315dm3kPa/molK)(298K) V = 1.25 dm3 CO2

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 15.0 L non-STP ? g GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = 8.315 dm3kPa/molK WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315dm3kPa/molK) (294K) n = 0.597 mol O2 Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT 

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