2C6H5Cl(chlorobenzene)+C2HOCl3(chloral)=C14H9Cl5(DDT)+H2O
in a gov. lab 1142 g of chlorobenzene is reacted with 485 g of chloral.
a). what mass of DDT is formed? 1166.179 g of C14H9Cl5
b). which reactant is limiting?C2HOCl3 which reactant is in excess?C6H5Cl
c). what mass of the excess reactant is left over? 397.56g C6H5Cl
d). if the actual yield is 200.0 g, what is the percent yield? 50.30%
a)I obtained 1166.5 which would round to 1166 g (but the 485 limits us to 3 sig figures).
b)right on both.
c)I obtained 740.77 g chlorobenzene used which gives 401.22 remaining. The difference may be in rounding. I left all the numbers in my calculator as I went from one step to the other. I think your number is close enough that you probably worked the problem correctly. Of course we would get slightly different numbers for percent yield; I used your numbers and 50.306%
which will round to 50.31 but we aren't allowed that many significant figures anyway. Good work.
To answer these questions, we need to apply stoichiometry and the concept of limiting reactants. Let's go through each question step by step:
a) To find the mass of DDT formed, we first need to determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed. To determine this, we need to calculate the moles of each reactant.
First, calculate the moles of chlorobenzene (C6H5Cl):
Molar mass of C6H5Cl = (12.01 g/mol * 6) + (1.01 g/mol * 5) + 35.45 g/mol = 112.56 g/mol
Moles of C6H5Cl = Mass of C6H5Cl / Molar mass of C6H5Cl
= 1142 g / 112.56 g/mol
= 10.14 mol
Next, calculate the moles of chloral (C2HOCl3):
Molar mass of C2HOCl3 = (12.01 g/mol * 2) + 1.01 g/mol + 35.45 g/mol + 3 * 16.00 g/mol = 147.43 g/mol
Moles of C2HOCl3 = Mass of C2HOCl3 / Molar mass of C2HOCl3
= 485 g / 147.43 g/mol
= 3.29 mol
Using the balanced equation, we can see that the stoichiometric ratio between chlorobenzene and DDT is 2:1. This means that for every 2 moles of chlorobenzene, 1 mole of DDT is formed. Since chlorobenzene is in excess, we only need to consider the moles of chloral to find the moles of DDT.
Moles of DDT = Moles of C2HOCl3 / 2
= 3.29 mol / 2
= 1.64 mol
Now we can calculate the mass of DDT formed:
Mass of DDT = Moles of DDT * Molar mass of DDT
= 1.64 mol * 354.48 g/mol (molar mass of C14H9Cl5)
≈ 579.827 g
Therefore, the mass of DDT formed is approximately 579.827 g.
b) We have already determined that the limiting reactant is chloral (C2HOCl3) and the reactant in excess is chlorobenzene (C6H5Cl).
c) To find the mass of the excess reactant left over, we first need to calculate the moles of DDT formed (which we already did in part a). Then we can calculate the moles of chlorobenzene that reacted with the moles of DDT.
Moles of chlorobenzene reacted = Moles of DDT * 2
= 1.64 mol * 2
= 3.28 mol
Now, we can calculate the moles of chlorobenzene left over:
Moles of chlorobenzene left over = Moles of chlorobenzene initially - Moles of chlorobenzene reacted
= 10.14 mol - 3.28 mol
= 6.86 mol
To find the mass of chlorobenzene left over, we multiply the moles by the molar mass:
Mass of chlorobenzene left over = Moles of chlorobenzene left over * Molar mass of chlorobenzene
= 6.86 mol * 112.56 g/mol (molar mass of C6H5Cl)
≈ 772.10 g
Therefore, the mass of the excess reactant (chlorobenzene) left over is approximately 772.10 g.
d) To calculate the percent yield, we need to compare the actual yield to the theoretical yield and then calculate the percentage.
Theoretical yield is the maximum amount of product that can be obtained based on stoichiometry calculations. In this case, it is the mass of DDT formed, which we found to be approximately 579.827 g in part a.
Percent yield = (Actual yield / Theoretical yield) x 100
= (200.0 g / 579.827 g) x 100
≈ 34.49%
Therefore, the percent yield is approximately 34.49%.
a). To find the mass of DDT formed, we will need to determine the limiting reactant and calculate the theoretical yield.
The balanced equation for the reaction is:
2C6H5Cl + C2HOCl3 → C14H9Cl5 + H2O
Given:
Mass of chlorobenzene (C6H5Cl) = 1142 g
Mass of chloral (C2HOCl3) = 485 g
Actual yield of DDT (C14H9Cl5) = 1166.179 g
First, we need to determine the limiting reactant. To do this, we compare the amounts of the reactants used to the stoichiometry of the balanced equation.
Molar mass of C6H5Cl = 112.56 g/mol
Molar mass of C2HOCl3 = 147.4 g/mol
Next, we calculate the number of moles for each reactant:
Moles of C6H5Cl = Mass / Molar mass = 1142 g / 112.56 g/mol = 10.14 mol
Moles of C2HOCl3 = Mass / Molar mass = 485 g / 147.4 g/mol = 3.29 mol
From the balanced equation, we can see that the ratio of moles between C6H5Cl and C2HOCl3 is 2:1. This means that every 1 mole of C2HOCl3 requires 2 moles of C6H5Cl. However, we have more moles of C6H5Cl (10.14 mol) compared to C2HOCl3 (3.29 mol). This indicates that C2HOCl3 is the limiting reactant.
Now we can calculate the theoretical yield of DDT using the moles of the limiting reactant:
Moles of C2HOCl3 reacted = 3.29 mol
Moles of C14H9Cl5 formed = 3.29 mol (according to stoichiometry)
Molar mass of C14H9Cl5 = 354.49 g/mol
Theoretical yield of DDT = Moles of DDT × Molar mass of DDT = 3.29 mol × 354.49 g/mol = 1164.42 g
Therefore, the mass of DDT formed is approximately 1164.42 g.
b). The limiting reactant is C2HOCl3 (chloral), and the reactant in excess is C6H5Cl (chlorobenzene).
c). To determine the mass of the excess reactant leftover, we need to find the amount of the limiting reactant consumed and subtract it from the initial mass of the reactant.
Moles of C2HOCl3 reacted = 3.29 mol
Moles of C6H5Cl required (according to the balanced equation) = 2 × 3.29 mol = 6.58 mol
Now, we can calculate the mass of C6H5Cl remaining:
Moles of C6H5Cl initially = 10.14 mol
Moles of C6H5Cl consumed = 6.58 mol
Moles of C6H5Cl remaining = 10.14 mol - 6.58 mol = 3.56 mol
Mass of C6H5Cl remaining = Moles of C6H5Cl remaining × Molar mass of C6H5Cl = 3.56 mol × 112.56 g/mol = 400.63 g
Therefore, the mass of the excess reactant (C6H5Cl) left over is approximately 400.63 g.
d). Percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.
Actual yield = 200.0 g
Theoretical yield = 1164.42 g
Percent yield = (Actual yield / Theoretical yield) × 100 = (200.0 g / 1164.42 g) × 100 = 17.17%
Therefore, the percent yield is approximately 17.17%.