The tetraethyl lead, Pb(C2H5)4 in a 25.00 mL sample of gasoline was shaken with 15.00 mL of  0.02095 M I2. The reaction is Pb(C2H5)4 + I2 → Pb(C2H5)3I + C2H5I  After the reaction was complete, the unused I2 was titrated with 6.09 mL of 0.03465 M Na2S2O3.  I2 + 2Na2S2O3 → 2NaI + Na2S4O6  Calculate the weight in milligrams of Pb(C2H5)4 (323.4 g/mol) in each milliliter of the gasoline. A. 8.100 mg/ml B. 5.400 mg/ml C. 2.700 mg/ml D.10.80 mg/ml

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