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
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
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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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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