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Chapter 3: Problem 114

DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction: $$ 2 \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl}+\mathrm{C}_{2} \mathrm{HOCl}_{3} \longrightarrow \mathrm{C}_{14} \mathrm{H}_{4} \mathrm{Cl}_{5}+\mathrm{H}_{2} \mathrm{O} $$ \(\begin{array}{ll}\text { orobenzenc chloral } & \mathrm{D}\end{array}\) In a government lab, \(1142 \mathrm{~g}\) of chlorobenzene is reacted with \(485 \mathrm{~g}\) of chloral. a. What mass of DDT is formed? b. Which reactant is limiting? Which is in excess? c. What mass of the excess reactant is left over? d. If the actual yield of DDT is \(200.0 \mathrm{~g}\), what is the percent yield?

Short Answer

Expert verified
a. The mass of DDT formed is 1165 g. b. Chloral (C2HOCl3) is the limiting reactant, and chlorobenzene (C6H3Cl) is the excess reactant. c. The mass of the excess reactant left over is 401.63 g. d. The percent yield is 17.17%.

Step by step solution

01

Write down and balance the chemical equation

Our given balanced chemical equation is: $$ 2 \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl}(\text{chlorobenzene})+\mathrm{C}_{2} \mathrm{HOCl}_{3}(\text{chloral}) \longrightarrow \mathrm{C}_{14} \mathrm{H}_{4} \mathrm{Cl}_{5}(\text{DDT})+\mathrm{H}_{2} \mathrm{O} $$
02

Convert mass of reactants to moles

To do this, we will divide the given mass by the molar mass of each reactant. Molar mass of chlorobenzene (C6H3Cl) = 112.5 g/mol Moles of chlorobenzene = 1142 g / 112.5 g/mol = 10.15 mol Molar mass of chloral (C2HOCl3) = 147.5 g/mol Moles of chloral = 485 g / 147.5 = 3.29 mol
03

Determine the limiting reactant

To determine the limiting reactant, we will use stoichiometry relating the moles of reactants. The given balanced equation: $$ 2 \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl} + \mathrm{C}_{2} \mathrm{HOCl}_{3} \longrightarrow \mathrm{C}_{14} \mathrm{H}_{4} \mathrm{Cl}_{5}+\mathrm{H}_{2}\mathrm{O} $$ We use mole ratios to see which reactant runs out first: 10.15 mol C6H3Cl / 2 (it's stoichiometric coefficient) = 5.075 3.29 mol C2HOCl3 / 1 = 3.29 Since 3.29 < 5.075, chloral (C2HOCl3) is the limiting reactant.
04

Calculate the mass of DDT produced

Using stoichiometry with the limiting reactant, we can calculate the mass of DDT produced. Molar mass of DDT (C14H4Cl5) = 354.5 g/mol Moles of DDT = moles of chloral * (1 mol DDT / 1 mol chloral) = 3.29 mol Mass of DDT = moles of DDT * molar mass of DDT = 3.29 * 354.5 = 1165 g
05

Determine mass of excess reactant left over

Now, we will find the mass of the excess reactant (chlorobenzene) left over. We start by calculating the amount of chlorobenzene that reacted with chloral: Moles of chlorobenzene that reacted = moles of chloral * (2 mol C6H3Cl / 1 mol C2HOCl3) = 3.29 * 2 = 6.58 mol Now subtract the moles of chlorobenzene that reacted from the initial moles of chlorobenzene: Moles of chlorobenzene left = 10.15 - 6.58 = 3.57 mol Mass of chlorobenzene left = moles of chlorobenzene left * molar mass of C6H3Cl = 3.57 * 112.5 = 401.63 g
06

Calculate the percent yield

Given that the actual yield of DDT is 200.0 g, we can calculate the percent yield. Percent yield = (actual yield / theoretical yield) * 100 = (200 / 1165) * 100 = 17.17% In summary: a. The mass of DDT formed is 1165 g. b. The limiting reactant is chloral (C2HOCl3) and the excess reactant is chlorobenzene (C6H3Cl). c. The mass of the excess reactant left over is 401.63 g. d. The percent yield is 17.17%.

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