I have a problem that says: 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 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 metal from the circuit board is:

Question

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

A plant needs to produce 5000 circuit boards, each with a surface area measuring 2.0 in X 3.0 in. The boards are covered with a 0.65 mm layer of copper metal. In subsequent processing, 85.0% of the copper is removed. Copper has a density of 8.96 g/cm3. Calculate the masses of Cu(NH3)4Cl2 and NH3 needed to produce the circuit boards, assuming the reaction used gives a 97% yield.

I started out by converting 2 and 3 in to cm, getting 5.08 and 7.62 cm respectively. Then I multiplied them and 0.65 nm together (coverted to cm) and got 2.5161 cm3. I multiplied the given density of 8.96 g/cm3 by the volume i just got, which gave me 2.254 x 10^-5 g. I multiplied that by 5,000 since there are 5,000 boards and that was just the amount of Cu per one board. That gave me 0.11272 g Cu. Since 85% of the copper is removed i multiplied that by 0.85 to get the amount of Cu removed and remaining, which were 0.095812 g and 0.016908 g respectively.

Now I am stuck and do not know how to find how much Cu(NH3)4Cl2 and NH3 needed to produce the boards. I don't know if anything I did above was even correct so if someone could help it would be greatly appreciated!!

Answer 1

edit: 2.5161 x 10^-6 cm

Answer 2

I don't agree with those numbers.
2.54*2*2.54*3*0.065*8.96 = 22.54 g/board and that x 5000 is 1.1127E5 g Cu.
That means you redo amount removed and remaining. Convert g Cu to mols. mols Cu = grams Cu/atomic mass Cu.

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

For every 1 mol Cu you will need 1 mol Cu(NH3)4Cl2 and 4 mols NH3. Convert mols Cu(NH3)2Cl2 and mols NH3 then to g. g = mols x molar mass or mols x atomic mass.
Finally since this yield is only 97%, then g Cu/0.97 = final g Cu needed and do the same kind of thing for g Cu(NH3)4Cl2

Answer 3

Ok I understand how to finish it, but if its 0.65 nanometers would my numbers be correct? Also, how do you know to divide g Cu by 0.97? I came across that last year but never really understood why you divide and not multiply. Thanks!