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:
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!!
edit: 2.5161 x 10^-6 cm
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
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!
You are on the right track! Let's break down the problem further to find the masses of Cu(NH3)4Cl2 and NH3 needed.
First, let's calculate the mass of copper (Cu) that remains on one circuit board after 85% of it is removed. You correctly calculated it as 0.016908 g.
Next, let's determine the moles of Cu on one circuit board. We can use the molar mass of Cu (63.55 g/mol) and apply the formula: moles = mass / molar mass.
So, the moles of Cu on one circuit board are:
moles of Cu = 0.016908 g / 63.55 g/mol ≈ 0.000266 mol.
Now, let's find the moles of Cu(NH3)4Cl formed in the reaction. By stoichiometry, the mole ratio between Cu and Cu(NH3)4Cl is 1:2. So the moles of Cu(NH3)4Cl2 formed are:
moles of Cu(NH3)4Cl2 = 2 * 0.000266 mol = 0.000532 mol.
Since the reaction is assumed to have a 97% yield, the moles of Cu(NH3)4Cl2 needed to produce 5000 circuit boards are:
moles of Cu(NH3)4Cl2 needed = 0.000532 mol / 0.97 ≈ 0.000549 mol.
Finally, let's find the masses of Cu(NH3)4Cl2 and NH3 needed.
The molar mass of Cu(NH3)4Cl2 is 214.06 g/mol. Therefore, the mass of Cu(NH3)4Cl2 needed is:
mass of Cu(NH3)4Cl2 = moles of Cu(NH3)4Cl2 needed * molar mass of Cu(NH3)4Cl2
= 0.000549 mol * 214.06 g/mol ≈ 0.117 g.
To find the mass of NH3 needed, we need to consider the stoichiometry of the reaction. From the balanced equation, we can see that it requires 4 moles of NH3 for every 1 mole of Cu(NH3)4Cl2. So, the moles of NH3 needed are:
moles of NH3 needed = 4 * moles of Cu(NH3)4Cl2 needed
= 4 * 0.000549 mol ≈ 0.002196 mol.
The molar mass of NH3 is 17.03 g/mol. Therefore, the mass of NH3 needed is:
mass of NH3 = moles of NH3 needed * molar mass of NH3
= 0.002196 mol * 17.03 g/mol ≈ 0.037 g.
To summarize, the masses of Cu(NH3)4Cl2 and NH3 needed to produce the 5000 circuit boards are approximately 0.117 g and 0.037 g, respectively.