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Unformatted text preview: Temperature of H 2 O: 27 ° C + 273.15= 300.15K Pressure in H 2 O: 26.74mmHg = 3.52atm Beaker with water (after reaction): 682.4g 259.6g of water displaced = 259.6g of N 2 produced P total = P H2O +P N 98.4atm = 3.52atm + P N P N = 94.9atm V= 259.6g/ 0.9965g/mL = 260.5mL= 0.2605L PV=nRT 94.9atm (0.2605L) = n(0.08206atm ⋅ L/mol ⋅ K)(300.15K) (24.72 atm ⋅ L)/(24.63atm ⋅ L/mol) = n n N = 1.00mol 259.6g N 2 (1mol/28.0134g N 2 )= 9.267mol N 2 Conclusion: The nitrite salt was determined to be . The accuracy of the experiment was determined by calculating a percent error of %error =  / =...
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This note was uploaded on 05/25/2008 for the course CHEM 2080 taught by Professor Davis,f during the Spring '07 term at Cornell.
 Spring '07
 DAVIS,F
 Chemistry, Stoichiometry

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