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 mass of copper removed from each board

First, we need to determine the volume of the copper removed from each board. We know that the copper layer is 0.60 mm thick and the dimensions of each board are \(8.0 \times 16.0 \mathrm{cm}\). To calculate the volume of the copper layer, multiply the area of the board by its thickness and convert the thickness from mm to cm. Area: \(8.0 \times 16.0 \mathrm{cm^{2}} = 128 \mathrm{cm^{2}}\) Thickness: \(0.60 \mathrm{mm} = 0.060 \mathrm{cm}\) Copper volume per board: \(128 \mathrm{cm^{2}} \times 0.060 \mathrm{cm} = 7.68 \mathrm{cm^{3}}\) Since 80% of the copper is removed, we can calculate the volume of copper removed: Volume of copper removed: \(0.80 \times 7.68 \mathrm{cm^{3}} = 6.144 \mathrm{cm^{3}}\) Now, we can find the mass of copper removed from each board by using the density of copper: Mass of copper removed: \(6.144 \mathrm{cm^{3}} \times 8.96 \mathrm{g/cm^{3}} = 55.055 \mathrm{g}\)

02

Calculate the total mass of copper removed from all boards

Multiply the mass of copper removed from each board by the number of boards: Total mass of copper removed: \(55.055 \mathrm{g} \times 10,000 = 550,550 \mathrm{g}\)

03

Use stoichiometry to find the required masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\)

From the given reaction: \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)\) For every mole of copper, one mole of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and four moles of \(\mathrm{NH}_{3}\) are required.

04

Calculate the moles of copper removed

Use the molar mass of copper (63.55 g/mol) to convert the total mass of copper removed into moles: Moles of copper removed: \(\frac{550,550 \mathrm{g}}{63.55 \mathrm{g/mol}} = 8,663.25 \mathrm{mol}\)

05

Calculate the required moles of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\)

For each mole of removed copper, we need one mole of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and four moles of \(\mathrm{NH}_{3}\): Moles of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) required: \(8,663.25 \mathrm{mol}\) Moles of \(\mathrm{NH}_{3}\) required: \(8,663.25 \mathrm{mol} \times 4 = 34,653 \mathrm{mol}\)

06

Calculate the required masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\)

Finally, we can use the molar masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) to convert the moles into masses: Mass of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) required: \(8,663.25 \mathrm{mol} \times (250.5 \mathrm{g/mol}) = 2,169,330.75 \mathrm{g}\) Mass of \(\mathrm{NH}_{3}\) required: \(34,653 \mathrm{mol} \times (17.03 \mathrm{g/mol}) = 590,570.59 \mathrm{g}\) So, the plant needs 2,169,330.75 g of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and 590,570.59 g of \(\mathrm{NH}_{3}\) to manufacture 10,000 printed circuit boards with 100% yield.

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