In the production of printed circuit boards for the electronics industry, a \(0.60\) -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 $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)$$ 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 \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) are needed to do this? Assume \(100 \%\) yield.

First, let's find out the volume of copper that needs to be removed from one board. Given, a \(0.60\)-mm layer of copper is laminated onto an insulating plastic board of dimensions \(8.0 \times 16.0 \mathrm{~cm}\). We also know that \(80 \%\) of the copper is removed from each board. Volume of copper on one board: \[V_\mathrm{Cu} = l \times w \times h\] Converting the layer thickness to cm and plugging in the values, we have: \[V_\mathrm{Cu} = 8.0 \times 16.0 \times 0.06\] Calculate the volume of copper that needs to be removed (\(80\%\)) from one board: \[V_\mathrm{CuRemoved} = 0.8 \times V_\mathrm{Cu}\]

Now that we have the volume of copper that needs to be removed from one board, we can calculate its mass. To do this, we can use the density of copper given as \(\rho_\mathrm{Cu} = 8.96 \mathrm{g}/\mathrm{cm}^3\). Mass of copper removed from one board: \[m_\mathrm{CuRemoved} = V_\mathrm{CuRemoved} \times \rho_\mathrm{Cu}\]

Now, let's find out the total mass of copper that needs to be removed from 10,000 boards. Total mass of copper removed from all boards: \[m_\mathrm{CuTotal} = 10{,}000 \times m_\mathrm{CuRemoved}\]

Now that we have the total mass of copper that needs to be removed from all the boards, we can use the stoichiometry of the given reaction to find the masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) needed. We have the balanced chemical reaction: \[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)\] First, calculate the moles of \(\mathrm{Cu}\) using the total mass of copper and its molar mass (\(63.55 \mathrm{g/mol}\)): \[n_\mathrm{Cu} = \frac{m_\mathrm{CuTotal}}{63.55}\] Now, we can find the moles of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) needed using stoichiometry: \[n_\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2} = \frac{1}{2}n_\mathrm{Cu}\] \[n_\mathrm{NH}_3 = 4\times n_\mathrm{Cu}\] Finally, find the masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) needed by multiplying the moles with their respective molar masses: \[m_\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}= n_\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2} \times ['Molar mass of \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}']\] \[m_\mathrm{NH}_3 = n_\mathrm{NH}_3\times[ 'Molar mass of \mathrm{NH}_3' ]\] This will give us the required masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) needed for the process.