5. The tetraethyl lead [Pb(C2H5)4] in a 25.00-mL sample of aviation gasoline was shaken with 15.00mL of 0.02095M I2. The reaction is: Pb(C2H5)4 + I2  Pb(C2H5)3I + C2H5I. After the reaction was complete, the unused I2 was titrated with 6.09mL of 0.03465M Na2S2O3. Calculate the weight in milligrams of Pb(C2H5)4 (323.4g/mol) in each liter of the gasoline.

Question

5.	The tetraethyl lead [Pb(C2H5)4] in a 25.00-mL sample of aviation gasoline was shaken with 15.00mL of 0.02095M I2. The reaction is: Pb(C2H5)4 + I2  Pb(C2H5)3I + C2H5I. After the reaction was complete, the unused I2 was titrated with 6.09mL of 0.03465M Na2S2O3. Calculate the weight in milligrams of Pb(C2H5)4 (323.4g/mol) in each liter of the gasoline.

DrBob222 · Answer

mols I2 initially = M x L = ?
Excess I2 titrated with S2O3^2- is
I2 + 2S2O3^2- ==> 2I^- + S4O6^2-
mols S2O3^2- = M x L = ?
Convert mols S2O3^2- to mols I2(that will be mols S2O3^2- x 1/2 = ? and this is the excess I2).
I2 used in the Pb(C2H5)4 = mols I2 initially - mols I2 excess = ?
Convert mols I2 used in the reaction to mols Pb(C2H5)4 using the coefficients in the balanced equation. That is 1 mol Pb(C2H5)4 = 1 mol I2.
Then g Pb(C2H5)4 = mols x molar mass and that is grams lead tetraethyl in 25.00 mL sample. Convert to mg, then convert mg/25 mL to mg/L. Post your work if you get stuck.