Chapter 5 Gases - 4 Gas Stoichiometry Dr. Sapna Gupta
Stoichiometry in Gases Amounts of gaseous reactants and products can be calculated by utilizing The ideal gas law to relate moles to T, P and V. Moles can be related to mass by the molar mass The coefficients in the balanced equation to relate moles of reactants and products Dr. Sapna Gupta/Gases-Stoichiometry
Solved Problem: When a.0-l bottle of concentrated HCl was spilled, 1. kg of CaCO 3 was required to neutralize the spill. What volume of CO was released by the neutralization at 735 mmhg and 0. C? First, write the balanced chemical equation: CaCO 3 (s) + HCl(aq) CaCl (aq) + H O(l) + CO (g) Second, calculate the moles of CO produced: Molar mass of CaCO 3 = 100.09 g/mol 1mol CaCO 1mol CO 3 3 1. 10 g CaCO3 100.09 g CaCO 3 1mol CaCO 3 = 11.99 mol V n = 11.99 mol P = 735 mmhg = 0.967 atm T = 0 C = 93 K L atm 11.99 mol 0.0806 (93 K) mol K (0.967 atm) V nrt P =.98 10 L (3 significant figures) Dr. Sapna Gupta/Gases-Stoichiometry 3
Collecting Gas Over Water Gases are often collected over water. The result is a mixture of the gas and water vapor. The total pressure is equal to the sum of the gas pressure and the vapor pressure of water. The partial pressure of water depends only on temperature and is known (Table 5.6). The pressure of the gas can then be found using Dalton s law of partial pressures. Dr. Sapna Gupta/Gases-Stoichiometry 4
P P P P H H H P P P H H O O 769 mmhg 16.5 mmhg P H 75.5 mmhg P H 753 mmhg (no decimal places) Dr. Sapna Gupta/Gases-Stoichiometry 5
Solved Problem: You prepare nitrogen gas by heating ammonium nitrite: NH 4 NO (s) N (g) + H O(l) If you collected the nitrogen over water at 3 C and 77 mmhg, how many liters of gas would you obtain from 5.68 g NH 4 NO? P = 77 mmhg P vapor = 1.1 mmhg P gas = 706 mmhg T = 3 C = 96 K Molar mass NH 4 NO = 64.06 g/mol 1mol NH NO 1mol N 4 5.68 g NH4NO = 0.08887 mol N 64.04 g NH 4 NO 1mol NH 4 NO n = 0.0887 mol V nrt P V L atm 0.0887 mol 0.0806 (96 K) mol K 1atm 706 mmhg 760 mmhg =.3 L of N (3 significant figures) Dr. Sapna Gupta/Gases-Stoichiometry 6
Solved Problem: Oxygen was produced and collected over water at ºC and a pressure of 754 torr. KClO 3 (s) KCl(s) + 3 O (g) 35 ml of gas were collected and the vapor pressure of water at ºC is 1 torr. Calculate the number of moles of O and the mass of KClO 3 decomposed. n P total = P O + P H O = P O + 1 torr = 754 torr P O = 754 torr 1 torr = 733 torr = 733 / 760 atm V = 35 ml = 0.35 L T = ºC + 73 = 95 K 733 atm0.35 L 760 1.910 mol O Latm 0.0806 (95 K) molk O KClO 3 (s) KCl(s) + 3 O (g) 1.910 mol O mol KClO 3 mol O 3 1.6 g KClO 3 1mol KClO 3 1.06 g KClO 3 Dr. Sapna Gupta/Gases-Stoichiometry 7
Speed of Gas Root mean square (rms) speed (u rms ) For two gases (1 and ) Effect of Temperature on Molecular Speed (1 st graph) Effect of Molar Mass on Molecular Speed ( nd graph) Dr. Sapna Gupta/Gases-Stoichiometry 8
Diffusion and Effusion Diffusion The process whereby a gas spreads out through another gas to occupy the space uniformly. Below NH 3 diffuses through air. The indicator paper tracks its progress. Effusion The process by which a gas flows through a small hole in a container. A pinprick in a balloon is one example of effusion. Dr. Sapna Gupta/Gases-Stoichiometry 9
Real Gases At high pressure the relationship between pressure and volume does not follow Boyle s law. This is illustrated on the graph below. At high pressure, some assumptions of the kinetic theory no longer hold true. At high pressure: 1. the volume of the gas molecule is not negligible.. the intermolecular forces are not negligible. Dr. Sapna Gupta/Gases-Stoichiometry 10
Van der Waal s Equation An equation that is similar to the ideal gas law, but which includes two constants, a and b, to account for deviations from ideal behavior. The term V becomes (V nb) to account for the space between molecules. The term P becomes (P + n a/v ) to account for attraction/repulsion between molecules. Values for a and b are different for different gases and can be found in Table 5.7 The ideal gas law becomes van der Waal s equation a and b have specific values for each gas Dr. Sapna Gupta/Gases-Stoichiometry 11
Key Points Gas stoichiometry Collecting gas over water Gas mixtures Molecular speed Diffusion and effusion Deviation from ideal behavior Factors causing deviation Van der Waal s equation Dr. Sapna Gupta/Gases-Stoichiometry 1