3. In the production of printed circuit boards for the electronics industry, a 0

Question

3. 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 in finally removed by solvents. One etching reaction is: Cu(NH3)4 NHCl2 + 4 NH3 + Cu arrow (NH3)4Cl A plant needs to manufacture 10,000 printed circuit boards, each 8.0 x 16.0 cm in area. An average of 80.0% of the copper is removed from each board (density of copper = 8.96 g/cm^3). What masses of Cu(NH3)4Cl2 and NH3 are needed to do this?

Explanation / Answer

Volume of each board = 8 x 16 x 0.06 cm^3 = 7.68 cm^3

Mass of Cu in each board = Volume * Density = 7.68 * 8.96 g/cm^3

Mass of Cu = 68.81 g

Mass of Copper removed per board = 80/100 * 68.81 = 55.05 g

Mass of Cu removed from 10,000 boards = 5.50 x 10^5 g

Mol of Cu removed = 5.50 x 10^5 / 63.5 = 8.67 x 10^3 mol

Cu(NH3)4Cl2 + 4NH3 + Cu --------> 2 Cu(NH3)4Cl

1 mol of Cu gives 1 mol of Cu(NH3)4Cl2

8.67 x 10^3 mol of Cu gives of 8.67 x 10^3 Cu(NH3)4Cl2

Mass of Cu(NH3)4Cl2 = 8.67 x 10^3 * 202.5 = 1.76 x 10^6 g

1 mol of Cu requires 4 mol of NH3

8.67 x 10^3 mol of Cu requires 8.67 x 10^3 * 4 = 3.47 x 10^4 mol

Mass of NH3 required = 3.47 x 10^4 * 17 = 5.9 x 10^5 g

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