Question

in the production of printed circuit boards for the electronics industry, a 0.60 - 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

mathrm{cu}left(mathrm{nh}_{3}
ight)_{4} mathrm{cl}_{2}(a q)+4 mathrm{nh}_{3}(a q)+mathrm{cu}(s)longrightarrow 2 mathrm{cu}left(mathrm{nh}_{3}
ight)_{4} mathrm{cl}(a q)

a plant needs to manufacture 10,000 printed circuit boards, each ( 9.0 \times 18.0 mathrm{~cm} ) in area. an average of ( 80.0 % ) of the copper is removed from each board (density of copper ( =8.90 mathrm{~g} / mathrm{cm}^{3} ). what masses of ( mathrm{cu}left(mathrm{nh}_{3}
ight)_{4} mathrm{cl}_{2} ) and ( mathrm{nh}_{3} ) are needed to do this? assume ( 100 % ) yield.

mass of ( mathrm{cu}left(mathrm{nh}_{3}
ight)_{4} mathrm{cl}_{2}= ) ( mathrm{g} )

mass of ( mathrm{nh}_{3}= ) ( mathrm{g} )

Explanation:

Step1: Calculate the volume of copper removed per board

The area of each board is \(A = 9.0\times18.0\space cm^{2}=162\space cm^{2}\), the thickness \(t = 0.60\times10^{- 1}\space cm\). The volume of copper per board \(V=A\times t\).
\(V = 162\times0.60\times10^{-1}\space cm^{3}=9.72\space cm^{3}\)

Step2: Calculate the mass of copper removed per board

Given the density of copper \(
ho = 8.90\space g/cm^{3}\). Using the formula \(m=
ho V\), the mass of copper per board \(m_{Cu, per\space board}=
ho V\)
\(m_{Cu, per\space board}=8.90\times9.72\space g = 86.508\space g\)

Step3: Calculate the total mass of copper removed

The number of boards \(n = 10000\) and the percentage of copper removed \(80\%\). The total mass of copper removed \(m_{Cu, total}=n\times m_{Cu, per\space board}\times0.80\)
\(m_{Cu, total}=10000\times86.508\times0.80\space g=692064\space g\)

Step4: Use stoichiometry to find the mass of \(\ce{Cu(NH3)4Cl2}\)

From the balanced equation \(Cu(NH_{3})_{4}Cl_{2}(aq)+4NH_{3}(aq)+Cu(s)
ightarrow2Cu(NH_{3})_{4}Cl(aq)\), the molar ratio of \(Cu\) to \(\ce{Cu(NH3)4Cl2}\) is \(1:1\).
The molar mass of \(Cu\) is \(M_{Cu}=63.55\space g/mol\), and the molar mass of \(\ce{Cu(NH3)4Cl2}\) is \(M_{Cu(NH_{3})_{4}Cl_{2}}=63.55+(14 + 3\times1)\times4+2\times35.45=202.55\space g/mol\)
The number of moles of \(Cu\), \(n_{Cu}=\frac{m_{Cu, total}}{M_{Cu}}=\frac{692064}{63.55}\space mol\)
The mass of \(\ce{Cu(NH3)4Cl2}\), \(m_{Cu(NH_{3})_{4}Cl_{2}}=\frac{m_{Cu, total}}{M_{Cu}}\times M_{Cu(NH_{3})_{4}Cl_{2}}\)
\(m_{Cu(NH_{3})_{4}Cl_{2}}=\frac{692064}{63.55}\times202.55\space g\approx2.20\times10^{6}\space g\)

Step5: Use stoichiometry to find the mass of \(\ce{NH3}\)

The molar ratio of \(Cu\) to \(\ce{NH3}\) is \(1:4\). The molar mass of \(\ce{NH3}\) is \(M_{NH_{3}}=14 + 3\times1=17\space g/mol\)
The mass of \(\ce{NH3}\), \(m_{NH_{3}}=\frac{m_{Cu, total}}{M_{Cu}}\times4\times M_{NH_{3}}\)
\(m_{NH_{3}}=\frac{692064}{63.55}\times4\times17\space g\approx7.40\times10^{5}\space g\)

Answer:

Mass of \(\ce{Cu(NH3)4Cl2}\): \(2.20\times 10^{6}\space g\)
Mass of \(\ce{NH3}\): \(7.40\times 10^{5}\space g\)