4. The tetraethyl lead [Pb(C2H5)4] in a 25.00-ml sample of aviation gasoline was shaken with 15.00mL of 0.02095 M 12 . The reaction is: Pb(C;Hs), + ½ → Pb(C,H;)gl + CzH,l After the reaction was complete, the unused 12 was titrated with 6.09mL of 0.03465 M Na2S203. Calculate the weight (in milligrams) of Pb(C2H5)4 (MM 323.4) in each liter of the gasoline. 25,0,2 + l2 → 21 + S,O,² Answer: 2.70 x 10° mg/L gasoline

4. The tetraethyl lead [Pb(C2H5)4] in a 25.00-ml sample of aviation gasoline was shaken with 15.00mL of 0.02095 M 12 . The reaction is: Pb(C;Hs), + ½ → Pb(C,H;)gl + CzH,l After the reaction was complete, the unused 12 was titrated with 6.09mL of 0.03465 M Na2S203. Calculate the weight (in milligrams) of Pb(C2H5)4 (MM 323.4) in each liter of the gasoline. 25,0,2 + l2 → 21 + S,O,² Answer: 2.70 x 10° mg/L gasoline

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter4: Reactions In Aqueous Solution

Section: Chapter Questions

Problem 61QAP

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The tetraethyl lead [Pb(C2H5 )4 ] in a 25.00-mL sample of aviation gasoline was shaken with 15.00mL of 0.02095 M I2 . The reaction is: Pb(C2H5 )4 + I2 → Pb(C2H5 )3 I + C2H5I After the reaction was complete, the unused I2 was titrated with 6.09mL of 0.03465 M Na2 S2O3. Calculate the weight (in milligrams) of Pb(C2H5 )4 (MM 323.4) in each liter of the gasoline. 2S2O3 2- + I2 → 2I- + S4O6 2- Answer: 2.70 x 103 mg/L gasoline
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