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 AgCl. a. Develop a general equation that relates the average value of n to the mass of a given mixture of PCBs and the mass of AgCl produced. b. A 0.1947-g sample of a commercial PCB yielded 0.4791 g of AgCl. What is the average value of n for this sample?

To find a general equation that relates the average value of n, mass of PCB mixture (m_PCB), and mass of AgCl produced (m_AgCl), we should first look at the chemical reaction. We will label the PCB species as C12H(10-n)Cl_n, and will define the molar masses of all species involved: M_PCB, M_Cl2, M_AgCl, and M_HCl. $$ \mathrm{C}_{12} \mathrm{H}_{10}+n \mathrm{Cl}_{2} \rightarrow \mathrm{C}_{12} \mathrm{H}_{10-n} \mathrm{Cl}_{n}+n \mathrm{HCl} $$ Notice that for every mole of Cl_n produced in the reaction, n moles of Cl2 are used and n moles of HCl are produced. The mass of the Chlorine in PCB mixture (m_Cl) is also released as AgCl, so we can write: $$ m_{Cl} = m_{AgCl} \times \frac{M_{Cl}}{M_{AgCl}} = m_{PCB} \times \frac{M_{Cl.n}}{M_{PCB}} $$ Where M_Cl.n is the molar mass of Chlorine atoms in one mole of PCB in the given mixture. Now we can rewrite this equation to find the M_Cl.n in terms of m_PCB, and m_AgCl: $$ M_{Cl.n} = \frac{m_{PCB} \times M_{Cl}}{m_{AgCl}} \times M_{AgCl} $$ Now, the average value of n can be found by dividing M_Cl.n by the molar mass of Chlorine: $$ n_{avg} = \frac{M_{Cl.n}}{M_{Cl}} $$ Combining the equations: $$ n_{avg} = \frac{m_{PCB} \times M_{Cl}}{m_{AgCl}} \times M_{AgCl} \times \frac{1}{M_{Cl}} $$