Gas Stoichiometry Moles  Liters of a Gas: Non-STP 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

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 = 1.26 mol CO2 Plug this into the Ideal Gas Law to find liters. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 1.26 mol T = 25°C = 298 K R = 8.315 dm3kPa/molK WORK: PV = nRT (103 kPa)V =(1mol)(8.315dm3kPa/molK)(298K) V = 1.26 dm3 CO2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

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

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

Zn (s) + 2 HCl (aq) ZnCl2(aq) + H2(g) Gas Stoichiometry Find vol. hydrogen gas made when 38.2 g zinc react w/excess hydrochloric acid. Pres. = 107.3 kPa; temp.= 88oC. Zn (s) + 2 HCl (aq) ZnCl2(aq) + H2(g) 38.2 g excess X L P = 107.3 kPa (13.1 L) T = 88oC 1 mol Zn 1 mol H2 22.4 L O2 x L H2 = 38.2 g Zn = 13.1 L H2 65.4 g Zn 1 mol Zn 1 mol H2 Zn H2 The relationship between the amounts of gases (in moles) and their volumes (in liters) in the ideal gas law is used to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known. • Relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. • The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. • In the lab, gases produced in a reaction are collected by the displacement of water from filled vessels — the amount of gas can be calculated from the volume of water displaced and the atmospheric pressure. Combined Gas Law At STP, we’d use 22.4 L per 1 mol, but we aren’t at STP. 1 mol Zn 1 mol H2 x mol H2 = 38.2 g Zn = 0.584 mol H2 65.4 g Zn 1 mol Zn 88oC + 273 = 361 K V = n R T P 0.584 mol (8.314 L.kPa/mol.K)(361 K) P V = n R T = = 16.3 L 107.3 kPa

Zn (s) + 2 HCl (aq) ZnCl2(aq) + H2(g) Gas Stoichiometry Find vol. hydrogen gas made when 38.2 g zinc react w/excess hydrochloric acid. Pres. = 107.3 kPa; temp.= 88oC. Zn (s) + 2 HCl (aq) ZnCl2(aq) + H2(g) 38.2 g excess X L P = 107.3 kPa (13.1 L) T = 88oC 1 mol Zn 1 mol H2 22.4 L O2 x L H2 = 38.2 g Zn = 13.1 L H2 65.4 g Zn 1 mol Zn 1 mol H2 Zn H2 The relationship between the amounts of gases (in moles) and their volumes (in liters) in the ideal gas law is used to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known. • Relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. • The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. • In the lab, gases produced in a reaction are collected by the displacement of water from filled vessels — the amount of gas can be calculated from the volume of water displaced and the atmospheric pressure. Combined Gas Law At STP, we’d use 22.4 L per 1 mol, but we aren’t at STP. P1 = T1 = V1 = P2 = T2 = V2 = 101.3 kPa P1 x V1 T1 P2 x V2 T2 (101.3 kPa) x (13.1 L) = (107.3 kPa) x (V2) 273 K = 273 K 361 K 13.1 L 107.3 kPa V2 = 16.3 L 88 oC + 273 = 361 K X L

2 Mg (s) + CO2 (g) 2 MgO (s) + C (s) What mass solid magnesium is required to react w/250 mL carbon dioxide at 1.5 atm and 77oC to produce solid magnesium oxide and solid carbon? 2 Mg (s) + CO2 (g) 2 MgO (s) + C (s) X g Mg 250 mL 0.25 L V = 250 mL 0.25 L oC + 273 = K T = 77oC 350 K P = 1.5 atm 151.95 kPa n = R T P V 151.95 kPa 1.5 atm (0.250 L) P V = n R T n = = 0.013 mol CO2 0.0821 L.atm / mol.K 8.314 L.kPa / mol.K (350 K) 2 mol Mg 24.3 g Mg x g Mg = 0.013 mol CO2 = 0.63 g Mg 1 mol CO2 1 mol Mg CO2 Mg

Gas Stoichiometry 2 Na + Cl2 NaCl 2 P1 x V1 T1 P2 x V2 T2 = How many liters of chlorine gas are needed to react with excess sodium metal to yield 5.0 g of sodium chloride when T = 25oC and P = 0.95 atm? 2 Na + Cl2 NaCl 2 excess X L 5 g 1 mol NaCl 1 mol Cl2 22.4 L Cl2 x g Cl2 = 5 g NaCl = 0.957 L Cl2 58.5 g NaCl 2 mol NaCl 1 mol Cl2 P1 x V1 T1 P2 x V2 T2 Ideal Gas Method = P1 = 1 atm T1 = 273 K V1 = 0.957 L P2 = 0.95 atm T2 = 25 oC + 273 = 298 K V2 = X L (1 atm) x (0.957 L) (0.95 atm) x (V2) = 273 K 298 K V2 = 1.04 L

Gas Stoichiometry 2 Na + Cl2 NaCl 2 V = n R T P P V = n R T V = 1.04 L How many liters of chlorine gas are needed to react with excess sodium metal to yield 5.0 g of sodium chloride when T = 25oC and P = 0.95 atm? 2 Na + Cl2 NaCl 2 excess X L 5 g 1 mol NaCl 1 mol Cl2 x g Cl2 = 5 g NaCl = 0.0427 mol Cl2 58.5 g NaCl 2 mol NaCl V = n R T P Ideal Gas Method P V = n R T P = 0.95 atm T = 25 oC + 273 = 298 K V = X L R = 0.0821 L.atm / mol.K n = 0.0427 mol 0.0427 mol (0.0821 L.atm / mol.K) (298 K) X L = 0.95 atm V = 1.04 L