ach, Second Edition Answer 124. DDT, an insecticide harmful to fish, birds, and

Question

ach, Second Edition Answer 124. DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction: chlorobenzenechloral DDT In a government lab, 1142 g of chlorobenzene is reacted with 485 g of chloral. a. What mass of DDT is formed, assuming 100% yield? b. Which seactant is umitingt Whlich is in excess? c. What mass of the excess reactant is left over? d. If the actual yield of DDT Hs 125. Bornite (Cu FeS,) is a copper ore used in the production of copper. When heated, the following reaetdion occurs o2 7 8 9 0 5 6

Explanation / Answer

2C6H5Cl   +   C2HOCl3   =   C14H9Cl5   +   H2O

In the above reaction equation,

2 moles of C6H5Cl reacts with 1 mole of C2HOCl3 to produce 1 mole of C14H9Cl5.

1142 g of C6H5Cl

Molar mass of C6H5Cl = 112.56 g/mol

So, 112.56 g of C6H5Cl = 1 mol

1 g of C6H5Cl = (1/112.56) mol

1142 g of C6H5Cl = (1142/112.56) mol
                         = 10.15 mol

485 g of C2HOCl3

Molar mass of C2HOCl3 = 147.4 g/mol

So, 147.4 g of C2HOCl3 = 1 mol

1 g of C2HOCl3 = (1/147.4) mol

485 g of C2HOCl3 = (485/147.4) mol
                              = 3.29 mol

Now, 3.29 x 2 = 6.58 < 10.15 mol

So, C2HOCl3 is the limiting reagent and C6H5Cl is the excess reagent.

(a)

Again;

2C6H5Cl   +   C2HOCl3   =   C14H9Cl5   +   H2O

In the above reaction equation,

1 mole of C2HOCl3 produces 1 mole of C14H9Cl5.

3.29 mole of C2HOCl3 produces 3.29 mole of C14H9Cl5.

Molar mass of C14H9Cl5 = 354.5 g/mol

So, 1 mole of C14H9Cl5 = 354.5 g

3.29 moles of C14H9Cl5 = 3.29 x 354.5 g
                                        = 1166.31 g

(c)

2C6H5Cl   +   C2HOCl3   =   C14H9Cl5   +   H2O

In the above reaction equation,

1 mole of C2HOCl3 reacts with 2 moles of C6H5Cl

3.29 moles of C2HOCl3 reacts with (2 x 3.29) moles of C6H5Cl

3.29 moles of C2HOCl3 reacts with 6.58 moles of C6H5Cl

So, excess moles of C6H5Cl left = 10.15 mol - 6.58 mol = 3.57 moles

Molar mass of C6H5Cl = 112.56 g/mol

So, 1 mol of C6H5Cl = 112.56 g

3.57 moles of C6H5Cl = 3.57 x 112.56 g = 401.84 g

(d)

Theoretical yield of C14H9Cl5 = 1166.31 g

Practical yield of C14H9Cl5 = 200 g

% Yield = (Practical yield / Theoretical yield) x 100
              = (200 / 1166.31) x 100
              = 17.15 %

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