In the production of printed circuit boards for the electronicsindustry, a 0.120

Question

In the production of printed circuit boards for the electronicsindustry, a 0.120 mm layer of copper is laminated onto aninsulating plastic board. Next a circuit pattern made of achemically resistant polymer is printed on the board. The unwantedcopper is removed by chemical etching and the protective polymer isfinally removed by solvents. One etching reaction is:

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


A plant needs to manufacture 8800 printed circuit boards, each7.60cm x 17.0cm in area. An average of 92.0% of the copper isremoved from each board (density ofcopper=8.96g/cm3.)
What mass of Cu(NH3)4Cl2 reagentis required?

Explanation / Answer

This is an interesting question. You will first need tocalculate the volume of copper that is laminated onto eachboard. The volume of copper is calculated by 7.60 cm * 17.0cm * 0.0120 cm = 1.5504 cm3. So, each board has1.5504 cm3 of copper on it. Using the density of copper, you know that 1.5504cm3 of copper equals (1.5504 cm3)(8.96 g/1cm3) = 13.89 g of copper. So, each board has 13.89g of copper on it. 92% of this copper will be involved in the reaction. That means that (0.92 * 13.89 g) = 12.78 g of copper will bereacted. All that's left is to realize that one mole ofCu(NH3)4Cl2 reacts with one moleof Cu and to get the molecular weights of these two compounds whichare 202.606 g/mol and 63.546 g/mol. Now, set up theequation: (12.78 g Cu)(1 mole Cu/63.546 g Cu)(1 molCu(NH3)4Cl2/1 mol Cu)(202.606 gCu(NH3)4Cl2/1 molCu(NH3)4Cl2) = 40.748 g Cu(NH3)4Cl2 So, a whopping 40.748 g ofCu(NH3)4Cl2 are required for eachboard. Lastly, the plant needs to make 8800 boards; so, theplant will nead(8800 * 40.748 g ofCu(NH3)4Cl2) = 358580.24 g ofCu(NH3)4Cl2 = 358.58024 kg ofCu(NH3)4Cl2. I hope this helps.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.