Stoichiometry and the Ideal Gas Law

Stoichiometry and the Ideal Gas Law - o C to get the volume...

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Stoichiometry and the Ideal Gas Law Lab Instructor: Dan Mack October 13, 2007
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Results and Discussion: The identity of an unknown nitrite salt was determined by reacting a known mass of it with sulfamic acid as follows: MNO 2 ( aq ) + HSO 3 NH 2 ( aq ) MHSO 4 ( aq ) + H 2 O( l ) + N 2 ( g ) The data obtained during the procedure is shown in Table 1 below: Table 1: Mass of empty 600mL beaker 219.6 g Mass of 600mL beaker and displaced H 2 O 564.7 g Mass of unkown salt 1.194 g Temperature of water in flask 30.8 o C Temperature of displaced water 22.9 o C Barometric Pressure 29.15 inHg To calculate the moles of MNO 2 used, the moles of N 2 were determined using the ideal gas law, PV = nRT The pressure of nitrogen gas was determined using Dalton’s law of partial pressures, P Total = P H 2 O + P N 2 29.15 inHg .03937in/mm = 21.07 mmHg + P N 2 P N 2 = 719 mmHg 719.3 mmHg 1 atm 760 mmHg = .9464 atm The mass of displaced water, 564.7 - 219.6 = 345.1 g H 2 O was divided by the density of water at 22.9
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Unformatted text preview: o C to get the volume of displaced water, and thus the volume of nitrogen gas produced: V = m r V = 345.1 g .99757 gmL-1 V = 345.9 mL = .3459 L The temperature of the nitrogen gas was assumed to be the same as that of the water in the flask. The number of moles of nitrogen, PV = nRT (.9464)(.3459) = n (.08206)(30.8 + 273) n = .01313 mol N 2 is equal to the number of moles of the unknown nitrate due to the molar ratio in the equation. The molar mass of the unknown compound, 1.194 g .01313 mol = 90.94 g/mol was used to determine the molar mass of the unknown metal by subtracting the molar mass of NO 2 from it. 90.94 – 46.01 = 44.93 g/mol Conclusion: Knowing the unknown metal must be an alkaline metal, the metal was determined to be potassium with an error of 14.91% | 44.93- 39.10 | 39.10 100% = 14.91%...
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