Copper is an excellent electrical conductor widely used in making electric circu
ID: 817445 • Letter: C
Question
Copper is an excellent electrical conductor widely used in making electric circuits. In producing a printed circuit board for the electronics industry, a layer of copper is laminated on a plastic board. A circuit pattern is then printed on the board using a chemically resistant polymer. The board is then exposed to a chemical bath that reacts with the exposed copper, leaving the desired copper circuit, which has been protected by the overlaying polymer. Finally, a solvent removes the polymer. One reaction used to remove the exposed copper from the circuit board is
Cu+Cu(NH3)4Cl2+4NH3=2Cu(NH3)4Cl
g/cm 3 a-
.Calculate the mass of Cu(NH3)4Cl2 needed to produce the circuit boards, assuming that the reaction used gives a 97% yield
b- Calculate the mass of NH3 needed to produce the circuit boards, assuming that the reaction used gives a 97% yield.
(BOTH IN KG NOT G)
PLEASE I NEED NUMBERS AS ANSWERS NOT JUST THE TO DO IT
Copper is an excellent electrical conductor widely used in making electric circuits. In producing a printed circuit board for the electronics industry, a layer of copper is laminated on a plastic board. A circuit pattern is then printed on the board using a chemically resistant polymer. The board is then exposed to a chemical bath that reacts with the exposed copper, leaving the desired copper circuit, which has been protected by the overlaying polymer. Finally, a solvent removes the polymer. One reaction used to remove the exposed copper from the circuit board is Cu+Cu(NH3)4Cl2+4NH3=2Cu(NH3)4Cl A plant needs to produce 4800 circuit boards, each with a surface area measuring 2.0in X 3.0inin. The boards are covered with a 0.61-mm layer of copper. In subsequent processing, 85% of the copper is removed. Copper has a density of 8.96 g/cm^3 Calculate the mass of Cu(NH3)4Cl2 needed to produce the circuit boards, assuming that the reaction used gives a 97% yield Calculate the mass of NH3 needed to produce the circuit boards, assuming that the reaction used gives a 97% yield.Explanation / Answer
mass of Cu(NH3)4Cl2 needed to produce the circuit boards= 0.059066 Kg.
mass of NH3 needed to produce the circuit boards= 0.0198392 Kg.
surface area = 2*3 = 6 in^2 = 6* 0.0254 ^ 2 = 0.00387096 m^2
volume of copper coated = surface area * thickness = 0.00387096 * 0.61 * 10 ^ -3 = 0.0000023612856 m^3
weight of copper coated = volume * density = 0.0000023612856 * 8.96 * 1000 = 0.021157118976 Kg.
amount of copper removed = 85 % of copper coated = 0.85 * 0.021157118976= 0.0179835511296 Kg
number of moles of copper removed = mass/ molecular weight = 0.0179835511296/63.546
number of moles of copper removed= 0.0179835511296/63.546 = 0.0002830005213483 Kilo moles of copper.
as the reaction yield is 97 % ,
let x be the number of Kilo moles of copper removed ,
then x*0.97 = 0.0002830005213483
x = 0.0002917531147920619 Kilo moles.
For 0.0002917531147920619 Kilo moles of copper , 0.0002917531147920619 Kilo moles of Cu(NH3)4Cl2 is required.
So, mass of Cu(NH3)4Cl2 = moles * molecular weight =0.0002917531147920619 * 202.452 = 0.059066 Kg.
mass of Cu(NH3)4Cl2 needed to produce the circuit boards= 0.059066 Kg.
number of moles of NH3 required = 4* x. = 4* x = 4* 0.0002917531147920619 = 0.0011670124591682476 Kilo moles of NH3.
So, mass of NH3 = moles * molecular weight = 0.0011670124591682476 * 17 = 0.0198392 Kg.
mass of NH3 needed to produce the circuit boards= 0.0198392 Kg.
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