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.
To calculate the weight of Pb(C2H5)4 in each liter of gasoline, you need to follow these steps:
Step 1: Calculate the moles of I2 reacted.
To do this, use the balanced equation: Pb(C2H5)4 + I2 → Pb(C2H5)3I + C2H5I
From the equation, we can see that the stoichiometric ratio between Pb(C2H5)4 and I2 is 1:1. This means that for every mole of Pb(C2H5)4 reacted with I2, there will be one mole of I2 reacted.
Since the volume and molarity of I2 are given, we can use the formula:
moles of I2 = volume of I2 (in L) × molarity of I2
moles of I2 = 15.00 mL × (0.02095 mol/L) = 0.03143 mol
Step 2: Calculate the moles of I2 reacted with Na2S2O3.
Again, using the stoichiometric ratio between I2 and Na2S2O3 from the balanced equation: I2 + 2Na2S2O3 → 2NaI + Na2S4O6
We see that the stoichiometric ratio between I2 and Na2S2O3 is 1:2. This means that for every mole of I2 reacted with Na2S2O3, there will be two moles of Na2S2O3 reacted.
Since the volume and molarity of Na2S2O3 are given, we can use the formula:
moles of Na2S2O3 = volume of Na2S2O3 (in L) × molarity of Na2S2O3
moles of Na2S2O3 = 6.09 mL × (0.03465 mol/L) = 0.02111 mol
Since the stoichiometric ratio between I2 and Na2S2O3 is 1:2, we can conclude that the moles of I2 reacted with Na2S2O3 is half of the moles of Na2S2O3.
moles of I2 reacted with Na2S2O3 = 0.02111 mol ÷ 2 = 0.01056 mol
Step 3: Calculate the moles of Pb(C2H5)4.
Since the stoichiometric ratio between Pb(C2H5)4 and I2 is 1:1, we can conclude that the moles of Pb(C2H5)4 reacted is equal to the moles of I2 reacted.
moles of Pb(C2H5)4 = 0.01056 mol
Step 4: Calculate the weight of Pb(C2H5)4.
To convert moles of Pb(C2H5)4 to grams, use the formula:
weight of Pb(C2H5)4 = moles of Pb(C2H5)4 × molar mass of Pb(C2H5)4
weight of Pb(C2H5)4 = 0.01056 mol × (323.4 g/mol) = 3.4156 g
Step 5: Calculate the weight of Pb(C2H5)4 in each liter of gasoline.
To convert grams of Pb(C2H5)4 to milligrams per liter of gasoline:
weight of Pb(C2H5)4 in mg/L = (3.4156 g / (25.00 mL / 1000)) × (1000 mL / 1 L) = 136.624 mg/L
So, the weight of Pb(C2H5)4 in each liter of gasoline is approximately 136.624 mg/L.
To calculate the weight of Pb(C2H5)4 in each liter of gasoline, we need to determine the number of moles of Pb(C2H5)4 in the given sample and then convert it to grams per liter.
Let's go step by step:
Step 1: Calculate the number of moles of I2 reacted
To do this, we will use the stoichiometry of the balanced equation. From the equation: Pb(C2H5)4 + I2 -> Pb(C2H5)3I + C2H5I, we can see that one mole of Pb(C2H5)4 reacts with one mole of I2.
The volume of I2 used is 15.00 mL, and the concentration of I2 is 0.02095 M.
Using the equation:
moles of I2 = volume (L) x concentration (M)
moles of I2 = (15.00 mL / 1000 mL/L) x 0.02095 M
Step 2: Calculate the number of moles of Pb(C2H5)4 reacted
Since the reaction is 1:1 between Pb(C2H5)4 and I2, the number of moles of Pb(C2H5)4 reacted is equal to the number of moles of I2 reacted.
Step 3: Calculate the number of moles of I2 remaining
To find the moles of I2 remaining, we need to subtract the moles of I2 reacted from the initial moles of I2 used. In this case, the initial moles of I2 used is also equal to the moles of I2 reacted.
Step 4: Calculate the number of moles of Na2S2O3 reacted
The volume of Na2S2O3 used is 6.09 mL, and the concentration of Na2S2O3 is 0.03465 M.
Using the equation:
moles of Na2S2O3 = volume (L) x concentration (M)
moles of Na2S2O3 = (6.09 mL / 1000 mL/L) x 0.03465 M
Step 5: Calculate the number of moles of I2 remaining after the Na2S2O3 titration
Since the reaction between I2 and Na2S2O3 is:
I2 + 2Na2S2O3 -> 2NaI + Na2S4O6,
we can see that 1 mole of I2 reacts with 2 moles of Na2S2O3. Therefore, the moles of I2 remaining can be calculated using the stoichiometry.
Step 6: Calculate the grams of Pb(C2H5)4 in each liter of gasoline
The molar mass of Pb(C2H5)4 is given as 323.4 g/mol.
To calculate the weight of Pb(C2H5)4 in each liter of gasoline, we need to convert the moles of Pb(C2H5)4 reacted to grams and then adjust for the sample volume.
Weight of Pb(C2H5)4 = (moles of Pb(C2H5)4) x (molar mass of Pb(C2H5)4)
Weight of Pb(C2H5)4 in each liter of gasoline = (Weight of Pb(C2H5)4 / 0.025 L)
Combining all the calculated values, we can now evaluate the weight in milligrams of Pb(C2H5)4 in each liter of the gasoline. Plug in the values calculated in each step to get the final 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.