Lab 4 - Stoichiometry and the Ideal Gas Law

Lab 4 - Stoichiometry and the Ideal Gas Law - The molar...

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Stoichiometry and the Ideal Gas Law October 15, 2007 1
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Results and Discussion: An unknown salt was mixed with sulfamic acid to release nitrogen gas. The resulting data was plugged into the PV=nRT to find out the identity of the salt. The procedure displaced 230ml of water, which is equivalent to the volume of nitrogen gas produced. P = 756.2mmHg/760; P = .995atm V= 230ml; V = .23 L T = 20 oC ; T = 293.15 oK R = .082057 n = unknown PV=nRT .995atm (.23 L) = n(.082057)(293.15 oK ) n = .009513 mol of nitrogen
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Unformatted text preview: The molar mass of the unknown metal M was determined by using molar ratios and the masses of the two other elements (oxygen and nitrogen) in the compound. .009513 mol of Nitrogen = .009513 mol of MNO 2 .009513 x .816g MNO 2 = 90.5g MNO 2 The molar weights of oxygen and nitrogen were subtracted. 90.5g MNO 2 – 14g N – 32g O = 44.5g M 44.5g = element Scandium Conclusion: The unknown metal was identified as Scandium 2...
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