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Gas Stoichiometry

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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 conversion Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

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1 mol CaCO 3 100.09g CaCO 3 Gas Stoichiometry Problem 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. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

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WORK: PV = nRT (103 kPa)V =(1mol)(8.315 dm 3 kPa/mol K )(298K) V = 1.26 dm 3 CO 2 Gas Stoichiometry Problem What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 1.26 mol T = 25°C = 298 K R = 8.315 dm 3 kPa/mol K Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

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WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315 dm 3 kPa/mol K ) (294K) n = 0.597 mol O 2 Gas Stoichiometry Problem 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 dm 3 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 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

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2 mol Al 2 O 3 3 mol O 2 Gas Stoichiometry Problem 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. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

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Find vol. hydrogen gas made when 38.2 g zinc react w /excess hydrochloric acid. Pres. = 107.3 kPa; temp.= 88 o C. Gas Stoichiometry 16.3 L At STP, we’d use 22.4 L per 1 mol, but we aren’t at STP. Zn (s) + 2 HCl (aq) ZnCl 2 (aq) + H 2 (g) 38.2 g excessX L P = 107.3 kPa T = 88 o C ZnH2H2 x L H 2 = 38.2 g Zn 65.4 g Zn = 13.1 L H 2 1 mol Zn 1 mol H 2 1 mol Zn 22.4 L O 2 1 mol H 2 (13.1 L) Combined Gas Law x mol H 2 = 38.2 g Zn 65.4 g Zn = 0.584 mol H 2 1 mol Zn 1 mol H 2 1 mol Zn P V = n R T V = n R T P = 0.584 mol (8.314 L. kPa/mol. K)(361 K) 107.3 kPa = 88 o C + 273 = 361 K

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P 2 x V 2 T 2 Find vol. hydrogen gas made when 38.2 g zinc react w /excess hydrochloric acid. Pres. = 107.3 kPa; temp.= 88 o C. Gas Stoichiometry 16.3 L At STP, we’d use 22.4 L per 1 mol, but we aren’t at STP. Zn (s) + 2 HCl (aq) ZnCl 2 (aq) + H 2 (g) 38.2 g excessX L P = 107.3 kPa T = 88 o C ZnH2H2 x L H 2 = 38.2 g Zn 65.4 g Zn = 13.1 L H 2 1 mol Zn 1 mol H 2 1 mol Zn 22.4 L O 2 1 mol H 2 (13.1 L) Combined Gas Law = P 1 = T 1 = V 1 = P 2 = T 2 = V 2 = = P 1 x V 1 T 1 (101.3 kPa) x (13.1 L) = (107.3 kPa) x (V 2 ) 273 K361 K V2V2 101.3 kPa 273 K 13.1 L 107.3 kPa 88 o C+ 273= 361 K X L

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(350 K) 151.95 kPa Mg (s) V = 250 mL What mass solid magnesium is required to react w /250 mL carbon dioxide at 1.5 atm and 77 o C to produce solid magnesium oxide and solid carbon? T = 77 o C P = 1.5 atm 250 mL X g Mg + CO 2 (g)MgO (s)+ C (s)22 0.25 L 350 K 151.95 kPa oC + 273 = K = 0.013 mol CO 2 P V = n R T n = R T P V n = 8.314 L. kPa / mol. K 0.0821 L. atm / mol. K 0.25 L (0.250 L) 1.5 atm CO 2 Mg x g Mg = 0.013 mol CO 2 1 mol CO 2 = 0.63 g Mg 2 mol Mg 24.3 g Mg 1 mol Mg

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Gas Stoichiometry How many liters of chlorine gas are needed to react with excess sodium metal to yield 5.0 g of sodium chloride when T = 25 o C and P = 0.95 atm? Na + Cl 2 NaCl 22 excessX L5 g x g Cl 2 = 5 g NaCl 1 mol NaCl 58.5 g NaCl2 mol NaCl 1 mol Cl 2 22.4 L Cl 2 1 mol Cl 2 = 0.957 L Cl 2 P 1 = 1 atm T 1 = 273 K V 1 = 0.957 L P 2 = 0.95 atm T 2 = 25 o C + 273 = 298 K V 2 = X L = P 2 x V 2 T 2 P 1 x V 1 T 1 (1 atm) x (0.957 L) (0.95 atm) x (V 2 ) 273 K298 K V 2 = 1.04 L = Ideal Gas Method

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Gas Stoichiometry How many liters of chlorine gas are needed to react with excess sodium metal to yield 5.0 g of sodium chloride when T = 25 o C and P = 0.95 atm? Na + Cl 2 NaCl 22 excessX L5 g x g Cl 2 = 5 g NaCl 1 mol NaCl 58.5 g NaCl2 mol NaCl 1 mol Cl 2 = 0.0427 mol Cl 2 P = 0.95 atm T = 25 o C + 273 = 298 K V = X L R = 0.0821 L. atm / mol. K n = 0.0427 mol P V = n R T 0.0427 mol (0.0821 L. atm / mol. K) (298 K) V = 1.04 L V = n R T P 0.95 atm Ideal Gas Method X L =

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