In the production of printed circuit boards for the electronics industry, a 0.26

Question

In the production of printed circuit boards for the electronics industry, a 0.260 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 6.10cm x 13.0cm in area. An average of 88.0% of the copper is removed from each board (density of copper=8.96g/cm3.) What mass of Cu(NH3)4Cl2 reagent is required?

Thanks

Explanation / Answer

volume of each board=(0.026cmx6.10cmx13cm)=2.0618cm^3

total volume=2.0618cm^3x8500=17525.3cm^3

percentage of Cu to be removed=88.0%

volume of Cu to be removed=17525.3cm^3x88/100=15422.264cm^3

converting volume to mass , we can use density given

mass of copper to be removed=volume xdensity=15422.264cm^3x8.96g/cm3=138183.485g

no of moles of copper=mass/molar mass=138183.485g/63.546 g/mol=2174.54mol

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

from the above equation it is clear that each mole of Cu requires 1 mole of Cu(NH3)4Cl2 reagent.so, 2174.54mol of Cu requires 2174.54mol of Cu(NH3)4Cl2 reagent.

mass of Cu(NH3)4Cl2 reagent required=no of moles of reagentxmolar mass of reagent

                                                          =2174.54mol x202.5741 g/mol=440506.012g

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