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Stoichiometry Lab

# Stoichiometry Lab - Results and Discussion The identity of...

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Results and Discussion The identity of an unknown nitrite salt (#6340) was found by dissolving the nitrite salt, which is a grainy white powder, in water and reacting it with excess sulfamic acid, a colorless solution with low viscosity. Dissolving the nitrite salt in water gave the solution a gelatinous appearance. Reacting the nitrite salt dissolved in water with the sulfamic acid resulted in rapid fizzing which gradually declined in rapidity as time progressed. It was known that the nitrite salt had the formula MNO 2 , in which M is an alkali metal cation. The reaction performed in the experiment was known to be: MNO 2 (aq) + HSO 3 NH 2 (aq) -> MHSO 4 (aq) + H 2 O (l) + N 2 (g) By determining the variables in the ideal gas law, PV = nRT, for the nitrogen gas produced, the number of moles of the nitrite salt could be calculated using the stoichiometric relationship between the number of moles of nitrogen gas product and nitrite salt reactant. The identity of the salt could then be determined using the mass of the nitrite salt and the number of moles of the nitrite salt. A mass of 2.102 g of unknown nitrite salt was reacted with 25.0 mL of sulfamic acid solution (80 g/ L solution), and the mass of water displaced by the nitrogen gas product was calculated, as follows: Mass of water displaced = mass of displaced water and beaker – mass of beaker Mass of water displaced = 659.3 g – 170.6 g Mass of water displaced = 488.7 g Due to the fact that the water was displaced by the production of nitrogen gas in the reaction of the nitrite salt and the sulfamic acid, the mass of the displaced water is assumed to be the mass of nitrogen gas produced. The ideal gas law includes volume as a variable. The volume of the nitrogen gas produced was indirectly determined by using the density formula to find the volume of water displaced, as follows: Density = mass / volume 1.00 g/mL = 488.7 g / volume Volume = 489 mL = 0.489 L A barometer was used to determine that the lab room had a pressure value of 757.5 mmHg. The

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