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.

Step by step solution

01

Calculate the volume of copper removed from one board

First, we need to find the total volume of copper removed from one board. The area of each board is \(8.0 \times 16.0 \mathrm{cm}^{2}\), and the thickness of the copper layer is 0.60 mm or 0.06 cm. So, the volume of copper on each board is: \(Area \times Thickness = 8.0 \times 16.0 \mathrm{cm}^{2} \times 0.06 \mathrm{cm} = 7.68 \mathrm{cm}^{3}\) Since 80% of the copper is removed, we find the volume of removed copper: \(7.68 \mathrm{cm}^{3} \times 0.80 = 6.144 \mathrm{cm}^{3}\)

02

Find the total volume of copper needed to be removed from 10000 boards

Now, we find the total volume of copper needed to be removed from 10000 boards by multiplying the volume of copper removed from one board by the number of boards (10000): \(6.144 \mathrm{cm}^{3/board} \times 10000 \mathrm{boards} = 61440 \mathrm{cm}^{3}\)

03

Calculate the mass of copper required for etching

We are given the density of copper, which is \(8.96 \mathrm{g} / \mathrm{cm}^{3}\). Using the density formula, we can find the mass of copper required for etching: \(Density = \frac{Mass}{Volume} \Rightarrow Mass = Density \times Volume\) \(Mass _{Cu} = 8.96 \mathrm{g/cm^{3}} \times 61440 \mathrm{cm}^{3} = 550758.4 \mathrm{g}\)

04

Use stoichiometry to determine the mass of Cu(NH3)4Cl2 and NH3

In the given reaction, one mole of Cu(NH3)4Cl2 reacts with 4 moles of NH3 to produce 2 moles of Cu from 1 mole of Cu(s). Using stoichiometry and the molar mass of Cu, Cu(NH3)4Cl2, and NH3, we can find the required masses for etching. Moles of Cu(s) needed: \(moles _{Cu} = \frac{550758.4 \mathrm{g}}{63.55 \mathrm{g/mol}} = 8669.73 \mathrm{moles}\) Moles of Cu(NH3)4Cl2 required: \(moles _{Cu(NH3)4Cl2} = \frac{1}{2} \times moles _{Cu} = 4334.86 \mathrm{moles}\) Mass of Cu(NH3)4Cl2 required: \(mass _{Cu(NH3)4Cl2} = 4334.86 \mathrm{moles} \times 228.78 \mathrm{g/mol}(molar\ mass) = 991818.49 \mathrm{g}\) Moles of NH3 required: \(moles _{NH3} = 4 \times moles _{Cu} = 34678.92 \mathrm{moles}\) Mass of NH3 required: \(mass _{NH3} = 34678.92 \mathrm{moles} \times 17.03 \mathrm{g/mol}(molar\ mass) = 590979.83 \mathrm{g}\) So, the plant needs 991818.49 g of Cu(NH3)4Cl2 and 590979.83 g of NH3 to manufacture 10000 printed circuit boards assuming 100% yield.

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