In the production of printed circuit boards for the electronics industry, a \(0.60-\mathrm{mm}\) layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s)\) \(\longrightarrow 2\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}(a q)\) a. Is this reaction an oxidation-reduction process? Explain. b. \(A\) plant needs to manufacture 10,000 printed circuit boards, each \(8.0 \times 16.0 \mathrm{~cm}\) in area. An average of \(80 . \%\) of the copper is removed from each board (density of copper \(=8.96\) \(\mathrm{g} / \mathrm{cm}^{3}\) ). What masses of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) are needed to do this? Assume \(100 \%\) yield.

To identify if it is an oxidation-reduction process, we can check the oxidation numbers of all elements. Remember that in a redox reaction, there will be a change in the oxidation number for at least one element in the reactants and products. The balanced chemical reaction is: \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s)\) \(\longrightarrow 2\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}(a q)\) Oxidation numbers: For \(\mathrm{Cl}\) in the reactants: -1 (as in all ionic compounds) For \(\mathrm{Cl}\) in the products: -1 (as in all ionic compounds) For \(\mathrm{N}\) in the reactants: -3 (in ammonia) For \(\mathrm{N}\) in the products: -3 (in ammonia) For \(\mathrm{Cu}\) in the reactants (Cu in complex and solid): 0 For \(\mathrm{Cu}\) in the products (Cu in complex): 0 As there are no changes in oxidation numbers, this reaction is not a redox process.

We are given that \(80\%\) of the copper layer is removed from each board. The area of each board is \(8.0 \times 16.0\,\mathrm{cm}^2\), and the copper layer is \(0.60\,\mathrm{mm}\) thick. We can calculate the volume of copper removed from each board and then multiply by the given density of copper to find the mass of copper to be removed. Volume of copper removed per board: \(= A \times t = (8.0\,\mathrm{cm} \times 16.0\,\mathrm{cm}) \times (0.60\,\mathrm{mm}) \times (80\%)\) \(= (8.0\,\mathrm{cm} \times 16.0\,\mathrm{cm}) \times (0.60\,\mathrm{mm}) \times 0.8\) Note that we need to convert 0.60 mm to cm: \(1\,\mathrm{mm} = 0.1\,\mathrm{cm}\) The volume of copper removed per board is: \(= (8.0\,\mathrm{cm} \times 16.0\,\mathrm{cm}) \times (0.06\,\mathrm{cm}) \times 0.8\) Now calculate the mass of copper removed per board: \(= \text{volume} \times \rho = (8.0\,\mathrm{cm} \times 16.0\,\mathrm{cm} \times 0.06\,\mathrm{cm} \times 0.8) \times 8.96\,\frac{\mathrm{g}}{\mathrm{cm}^3}\) Finally, calculate the mass of copper removed for 10,000 boards: \(= (8.0 \times 16.0 \times 0.06 \times 0.8 \times 8.96) \times 10{,}000\)

To determine the mass of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) required, we first need to calculate moles of copper removed. Calculate the moles of copper removed using the molar mass of Cu (63.5 g/mol): \(n_{\mathrm{Cu}} = \frac{\text{mass of Cu removed}}{\text{molar mass of Cu}}\)

Now, we can determine the moles of each reactant required by observing the coefficients in the balanced chemical equation. For \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\): From the balanced equation, 1 mole of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\) reacts with 1 mole of \(\mathrm{Cu}\). Therefore, we need the same moles of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\) as moles of \(\mathrm{Cu}\) removed. For \(\mathrm{NH}_{3}\): From the balanced equation, 4 moles of \(\mathrm{NH}_{3}\) reacts with 1 mole of \(\mathrm{Cu}\). Therefore, we need 4 times the moles of \(\mathrm{NH}_{3}\) as moles of \(\mathrm{Cu}\) removed.

Now that we know the moles of each reactant required, we can calculate their masses. For \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\): Molar mass of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\) = 253.0 g/mol. Mass of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\) required: \(= \text{moles of} \, \left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2} \times \text{molar mass of} \, \left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\) For \(\mathrm{NH}_{3}\): Molar mass of \(\mathrm{NH}_{3}\) = 17.0 g/mol. Mass of \(\mathrm{NH}_{3}\) required: \(= \text{moles of} \, \mathrm{NH}_{3} \times \text{molar mass of} \, \mathrm{NH}_{3}\) Now we have the masses of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) required for the production of 10,000 printed circuit boards.