Answer:
a. To determine the mass of DDT formed, we need to compare the reactants' mole ratios and convert the given masses into moles. The balanced equation tells us that the ratio between chlorobenzene (C6H5Cl) and DDT (C14HCl5) is 2:1. Similarly, the ratio between chloral (C2HOCl3) and DDT is 1:1.
First, let's find the moles of chlorobenzene:
Molar mass of chlorobenzene (C6H5Cl) = 112.56 g/mol
Moles of chlorobenzene = 1164 g / 112.56 g/mol = 10.35 mol
Next, let's find the moles of chloral:
Molar mass of chloral (C2HOCl3) = 147.39 g/mol
Moles of chloral = 474 g / 147.39 g/mol = 3.22 mol
Since the mole ratio between chlorobenzene and DDT is 2:1, and the moles of chlorobenzene are 10.35, the moles of DDT formed would be half that:
Moles of DDT formed = 10.35 mol / 2 = 5.18 mol
Now, let's calculate the mass of DDT:
Molar mass of DDT (C14HCl5) = 354.49 g/mol
Mass of DDT formed = 5.18 mol x 354.49 g/mol = 1834.02 g
Therefore, the mass of DDT formed, assuming 100% yield, is 1834.02 g.
b. To determine the limiting reactant and the excess reactant, we need to compare the moles of each reactant to their stoichiometric ratios.
For chlorobenzene (C6H5Cl):
Moles of chlorobenzene = 10.35 mol (as calculated in part a)
For chloral (C2HOCl3):
Moles of chloral = 3.22 mol (as calculated in part a)
The balanced equation shows that the ratio of chlorobenzene to chloral is 2:1. Since the moles of chloral are less than half the moles of chlorobenzene, chloral is the limiting reactant.
Therefore, chlorobenzene is in excess.
c. To find the mass of the excess reactant left over, we need to calculate the moles of chlorobenzene consumed in the reaction (assuming it is completely consumed) and then convert it to mass.
Moles of chlorobenzene consumed = (moles of chloral) x (moles of chlorobenzene in balanced equation / moles of chloral in balanced equation)
Moles of chlorobenzene consumed = 3.22 mol x (2 mol/1 mol) = 6.44 mol
Now, let's calculate the mass of the excess chlorobenzene:
Molar mass of chlorobenzene (C6H5Cl) = 112.56 g/mol
Mass of excess chlorobenzene = (moles of chlorobenzene - moles of chlorobenzene consumed) x molar mass of chlorobenzene
Mass of excess chlorobenzene = (10.35 mol - 6.44 mol) x 112.56 g/mol = 438.56 g
Therefore, the mass of the excess reactant (chlorobenzene) left over is 438.56 g.
d. To calculate the percent yield, we need to compare the actual yield (216.0 g) with the theoretical yield (as calculated in part a, 1834.02 g).
Percent yield = (actual yield / theoretical yield) x 100
Percent yield = (216.0 g / 1834.02 g) x 100 = 11.78%
Therefore, the percent yield is 11.78%.