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

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

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;H5), + I2 → Pb(C,H5)3l + C,H,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,- + 1, → 21 + S,0,²
Answer: 2.70 x 10° mg/L gasoline

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