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Chapter 4: Problem 116

Polychlorinated biphenyls (PCBs) have been used extensively as dielectric materials in electrical transformers. Because PCBs have been shown to be potentially harmful, analysis for their presence in the environment has become very important. PCBs are manufactured according to the following generic reaction: $$ \mathrm{C}_{12} \mathrm{H}_{10}+n \mathrm{Cl}_{2} \rightarrow \mathrm{C}_{12} \mathrm{H}_{10-n} \mathrm{Cl}_{n}+n \mathrm{HCl} $$ This reaction results in a mixture of \(\mathrm{PCB}\) products. The mixture is analyzed by decomposing the PCBs and then precipitating the resulting \(\mathrm{Cl}^{-}\) as \(\mathrm{AgCl}\) a. Develop a general equation that relates the average value of \(n\) to the mass of a given mixture of \(\mathrm{PCBs}\) and the mass of AgCl produced b. A \(0.1947-\mathrm{g}\) sample of a commercial \(\mathrm{PCB}\) yielded \(0.4791 \mathrm{~g}\) of \(\mathrm{AgCl}\). What is the average value of \(n\) for this sample?

Short Answer

Expert verified
The average value of n for the given commercial PCB sample is approximately 3.27.

Step by step solution

01

Write the balanced chemical equations

First we write the balanced chemical equations for formation of PCB and the decomposition of PCBs to form AgCl. Formation of PCB: \(C_{12}H_{10} + nCl_2 → C_{12}H_{10-n}Cl_n + nHCl\) Decomposition of PCBs and AgCl formation: \(C_{12}H_{10-n}Cl_n + nAgNO_3 → C_{12}H_{10-n}NO_3 + nAgCl\)
02

Develop a general equation that relates the average value of n, mass of PCBs, and mass of AgCl produced

We know that the number of moles is related to mass by the molar mass (MM) of each substance. Let the mass of PCBs used be m1 and mass of AgCl produced be m2. Moles of PCBs used = Moles of PCBs decomposed \(\frac{m1}{MM_{PCB}} = \frac{m2}{MM_{AgCl}}\) MM of PCB = \(12(12) + 1(10-n) + 35.5n\) (as \(nCl_2\) was used) MM of AgCl = \(107.9 + 35.5\) Now, substitute MM of PCB and MM of AgCl in the moles equation: \(\frac{m1}{144 + 10 - n + 35.5n} = \frac{m2}{143.4}\) This is the general equation relating the average value of n, mass of PCBs, and mass of AgCl produced.
03

Plug in the given values to calculate the average value of n

Given, mass of commercial PCB sample, m1 = 0.1947 g and mass of AgCl produced, m2 = 0.4791 g. \(\frac{0.1947}{144 + 10 - n + 35.5n} = \frac{0.4791}{143.4}\) Now, solve for n: \(n = \frac{143.4(0.1947) - 154(0.4791)}{35.5(0.4791)-(0.1947)}\) \(n = 3.27\) The average value of n for the given commercial PCB sample is approximately 3.27.

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Most popular questions from this chapter

A \(2.20-\mathrm{g}\) sample of an unknown acid (empirical formula \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}\) ) is dissolved in \(1.0 \mathrm{~L}\) of water. A titration required \(25.0\) \(\mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?

Give an example how each of the following insoluble ionic compounds could be produced using a precipitation reaction. Write the balanced formula equation for each reaction. a. \(\mathrm{Fe}(\mathrm{OH})_{3}(s)\) b. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\) c. \(\mathrm{PbSO}_{4}(s)\) d. \(\mathrm{BaCrO}_{4}(s)\)

Many oxidation-reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized. a. \(\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{CH}_{4}(g)+\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) c. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\)

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A stock solution containing \(\mathrm{Mn}^{2+}\) ions was prepared by dissolving \(1.584 \mathrm{~g}\) pure manganese metal in nitric acid and diluting to a final volume of \(1.000 \mathrm{~L}\). The following solutions were then prepared by dilution: For solution A. \(50.00 \mathrm{~mL}\) of stock solution was diluted to \(1000.0 \mathrm{~mL}\) For solution \(B, 10.00 \mathrm{~mL}\) of solution \(A\) was diluted to \(250.0 \mathrm{~mL}\). For solution \(C, 10.00 \mathrm{~mL}\) of solution \(B\) was diluted to \(500.0 \mathrm{~mL}\). Calculate the concentrations of the stock solution and solutions \(A, B\), and \(C\).
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