In the production of printed circuit boards for the electronics industry, a 0.71

Question

In the production of printed circuit boards for the electronics industry, a 0.715 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

Cu(NH3)4Cl2(aq) + 4NH3(aq) + Cu(s) ? 2Cu(NH3)4Cl(aq)

A plant needs to manufacture 8500 printed circuit boards, each 5.65cm x 11.0cm in area. An average of 85.0% of the copper is removed from each board (density of copper=8.96g/cm3.) What mass of Cu(NH3)4Cl2 reagent is required?

Explanation / Answer

First you need to determine how much copper is used in the final product after the etching is done. Determine volume of copper used to put on board: 0.60 mm x 80 mm x 160 mm = 7680 mm^3 of copper Next determine how much copper is removed after etching. 7680 x 0.8 = 6144 mm^3 removed after etching Next determine how much copper is left on the board after etching: 7680 mm^3 - 6144 mm^3 = 1536 mm^3 1536 mm^3 is the volume of copper needed for one board. Next multiply the amount of copper needed for one board by the total number of boards being produced: 1536 mm^3 x 10000 = 1.536 x 10^7 mm^3 of copper needed Determine the mass of the copper needed from the density formula d=m/v m=vd m=(1.536 x 10^7 mm^3 ) (8.96 g/cm^3) (1cm^3/1000 mm^3) m =137625.6 g of Cu needed Convert mass to moles 137625.6 g x (1mole Cu/63.54 g) = 2165.97 mole Cu Ratios from equation to determine mass of wanted compositions: 2165.97 mole Cu x (4 mole NH3/1 mole Cu) x 17.03g/1 mole NH3) = 68.12 g NH3 2165.97 mole Cu x (1 mole Cu(NH3)4Cl2 / 1 mole Cu) x 202.56 g / 1 mole Cu(NH3)4Cl2 = 438738.88 g Cu(NH3)4Cl2

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