Hydrogen Collection Over Water

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Presentation transcript:

Hydrogen Collection Over Water 14.23 mL of H2 gas is produced at 23.5C by the following reaction. If atmospheric pressure is 748.2 torr, how much Zn(s) is consumed? Reduction Oxidation 2 HCl(aq) + Zn(s)  ZnCl2(aq) + H2(g) 65.39 T = _____ 23.5C H2(g) gramsZn H2O(g) V = _____ 14.23 mL MolesZn P = _____ ? n = _____ Problem: The pressure in the collection tube is due to both H2O(g) and H2(g) HCl(aq) We need the pressure of H2(g) Zn(s)

Dalton’s Law of Partial Pressures Total Pressure H2 Partial Pressure H2O Partial Pressure = Ptot = PH2 + PH2O i.e. the total pressure is also ATMOSPHERIC PRESSURE Patm = PH2 + PH2O Rearrange..... PH2 = Patm - PH2O 748.2 torr (barometer) More Info Required

Partial Pressure of H2O @ 23.5 C (Provided on Exams) T= 23C PH2O = 21.1 torr T= 24C PH2O = 22.4 torr T= 23.5C PH2O = 21.75 torr P = 1.3 torr 1.3 torr 2 = 0.65 torr PH2O = 21.1 torr + 0.65 torr = 21.75 torr

Dalton’s Law of Partial Pressures Total Pressure H2 Partial Pressure H2O Partial Pressure Ptot = PH2 + PH2O i.e. the total pressure is the sum of all non-reactive gas partial pressures. Total pressure = atmospheric pressure Patm = PH2 + PH2O PH2 = Patm - PH2O - 748.2 torr (barometer) 21.75 torr More Info Required PH2 = 726.45 torr

Hydrogen Collection Over Water 65.39 2 HCl(aq) + Zn(s)  ZnCl2(aq) + H2(g) T = _____ 23.5C H2(g) gramsZn H2O(g) V = _____ 14.23 mL MolesZn P = _____ 726.45 torr n = _____ Problem: The pressure in the collection tube is due to both H2O(g) and H2(g) HCl(aq) We need the pressure of H2(g) Zn(s)

Hydrogen Collection Over Water 65.39 2 HCl(aq) + Zn(s)  ZnCl2(aq) + H2(g) T = _____ 23.5C 296.65 K ....... H2(g) 3.664  10-2 gramsZn H2O(g) V = _____ 0.01423 L....... 14.23 mL 5.58769 10-4 MolesZn 0.955855 atm....... P = _____ 726.45 torr n = _____ 5.59769  10-4 mols 1 mol Zn: 1 mol H2 n = P V R T (0.955855 atm) (0.01423 L) = (0.0820578 (L atm)/(mol K)) (296.65 K) HCl(aq) n = 5.58769  10-4 moles H2 Zn(s)