Question

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.

Answer 1

Weight in milligrams of Pb(C2H5)4 in each milliliter of the gasoline;

$=2.6×10 ^{3}mg$