In the production of printed circuit boards for the electronics industry, a 0.33

Question

In the production of printed circuit boards for the electronics industry, a 0.330 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 8200 printed circuit boards, each 6.55cm x 11.0cm in area. An average of 70.5% 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

Given the thickness of copper layer = 0.330 mm = (0.330 mm) x (1 cm / 10 mm) = 0.0330 cm

Area of the circuit board = 6.55cm x 11.0cm

Hence volume of copper in a single circiut board = Area x thickness

= 6.55cm x 11.0cm x 0.0330 cm

Now the volume of copper in 8200 circiut boards = 8200 x 6.55cm x 11.0cm x 0.0330 cm =  19500 cm3

Given the density of copper = 8.96g/cm3

Hence total mass of copper = Volume x density =  19500 cm3 x  8.96g/cm3  = 174720 g

given that 70.5% of the copper is removed from each board

Hence total mass of copper removed by etching = 174720 g x (70.5 / 100) = 123178 g

Atomic mass of copper = 63.5 g/mol

Hence total moles of copper removed by etching = 123178 g Cu x ( 1mol Cu / 63.5 g Cu)

= 1940 mol Cu

The balanced chemical reaction involved in the itching of copper is

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

1 mol 4 mol 1 mol 2 mol

In the abobe balanced chemical reaction, 1 mole of copper reacts with 1 mole of Cu(NH3)4Cl2(aq).

Hence 1940 moles of copper that will react with the moles of Cu(NH3)4Cl2(aq) =

= 1940 mol Cu x (1 mol Cu(NH3)4Cl2(aq) / 1 mol Cu) = 1940 mol Cu(NH3)4Cl2(aq)

Molecular mass of Cu(NH3)4Cl2(aq) = 202.5 g / mol

Hence the mass of   Cu(NH3)4Cl2(aq) equivalent to 1940 mol  Cu(NH3)4Cl2(aq)

= 1940 mol  Cu(NH3)4Cl2(aq) x (202.5 g  Cu(NH3)4Cl2(aq) / 1 mol  Cu(NH3)4Cl2(aq))

= 392800 g  Cu(NH3)4Cl2(aq)

or 392.8 Kg Cu (answer)

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