Question

The tetraethyl lead Pb(C2H5)4 in a 25.00 mL sample of aviation gasoline was shaken with 15.00 mL of 0.02263 M I2. (Hint: one mole of I2 reacts with one mole of the tetraethyl lead). After the reaction was complete, the unused I2 was titrated with 8.74 mL of 0.03949 M thiosulfate solution. Calculate the weight of Pb(C2H5)4 (MW = 323.4 g/mol) in each liter of gasoline.

Answer 1

mass tetraethyl lead in 1 Liter gasoline = 2.159 grams

Explanation

molarity I₂ = 0.02263 M

volume I₂ = 15.00 ml = 15.00 x 10^-3 L

moles I₂ = (molarity I₂) * (volume I₂)

moles I₂ = (0.02263 M) * (15.00 x 10^-3 L)

moles I₂ = 3.3945 x 10^-4 mol

moles thiosulfate = (molarity thiosulfate) * (volume thiosulfate)

moles thiosulfate = (0.03949 M) * (8.74 x 10^-3 L)

moles thiosulfate = 3.4514 x 10^-4 mol

moles I₂ reacted with thiosulfate = (1/2) * (moles thiosulfate)

moles I₂ reacted with thiosulfate = (0.5) * (3.4514 x 10^-4 mol)

moles I₂ reacted with thiosulfate = 1.7257 x 10^-4 mol

moles I₂ reacted with tetraethyl lead = (moles I₂) - (moles I₂ reacted with thiosulfate)

moles I₂ reacted with tetraethyl lead = (3.3945 x 10^-4 mol) - (1.7257 x 10^-4 mol)

moles I₂ reacted with tetraethyl lead = 1.669 x 10^-4 mol

moles tetraethyl lead = moles I₂ reacted with tetraethyl lead

moles tetraethyl lead = 1.669 x 10^-4 mol

mass tetraethyl lead = (moles tetraethyl lead) * (molar mass tetraethyl lead)

mass tetraethyl lead = (1.669 x 10^-4 mol) * (323.4 g/mol)

mass tetraethyl lead = 0.05397 g

mass tetraethyl lead in 1 Liter gasoline = (0.05397 g) * (1000 mL / 25.00 mL)

mass tetraethyl lead in 1 Liter gasoline = 2.159 g