CHEM
Lab #4 - Stoichiometry and the ideal gas law

# Lab #4 - Stoichiometry and the ideal gas law - Results and...

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Results and Discussion: An unidentified nitrate salt of the general formula MNO 2, where M + is an alkali metal cation(Li + , Na + , K + , Rb + ) was reacted with sulfamic acid, HSO 3 NH 2. Using the mass of the MNO 2 sample and the stoichiometric relationship between the moles of nitrogen gas and MNO 2, the identity of the nitrate salt was determined. 1. The MNO 2 was reacted with the HSO 3 NH 2 according to the reaction… MNO 2 (aq) + HSO 3 NH 2 (aq) MHSO 4 (aq) + H 2 O (l) + N 2 (g) The aqueous nitrate salt was reacted slowly with the sulfamic acid. As the nitrogen gas was produced the solution bubbled and water was pushed into the 600ml beaker. Mass of displaced water = mass of water and beaker – mass of beaker = 510.7g - 216.1g Mass of displaced water = 294.6g D = M/V 1.00g/ml = 294.6g/V V = 294.6ml Volume of nitrogen gas = volume of displaced water in beaker = 294.6ml = 0.2946 l 2. The temperature of the nitrogen gas was equal to the water temperature in the Erlenmeyer flask. Temperature = 35.8ºC =

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Unformatted text preview: 308.95 K 3. The pressure of the N 2 gas was found using Dalton’s Law of Partial Pressure. The total pressure was equal to the barometric pressure. The water vapor pressure is a function of temperature and was obtained using the chart. Pressure of H 2 O at 35.8ºC = 44.2824mmHg P Total = PH 2 O + PN 2 755 mmHg = 44.284 mmHg + PN 2 Pressure N 2 = 710.716 mmHg = 0.935 atm With the values for pressure, volume, and temperature, the moles of N 2 can be found. PV = nRT (0.935atm)(0.2946l) = n (0.08206)(308.95K) n = 0.01086 moles of nitrogen gas (0.01086moles nitrogen gas (1 mole MNO 2 ) = 0.01086 moles MNO 2 (1mole mole N 2) Mass of MNO 2 = (mass of vial + crystals) – mass of vial Mass of MNO 2 = 14.328g – 13.219 g = 1.109 g Molar mass of MNO 2 = mass of salt = 1.109g = 102.118g/mole Moles of salt 0.01086 moles Mass of NO 2 = N – 14.01 O - 16.00 x 2 = 32.00 46.00 g Mass of M = 102.118 – 46.00g =56g....
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