Question

n the 1990’s, the Boeing Co. developed a process for treating wastewater from electroplating and printed circuit board manufacturing. The central feature of the process was addition of scrap aluminum metal to an acidic solution containing dissolved copper ions. If the solution contained enough fluoride or chloride ions, a reaction proceeds in which the aluminum dissolved and the copper precipitated as metallic Cu(s), which could then be removed from the suspension by settling and/or filtration. The relevant reaction can be written as follows:233Cu2Al()3Cu()2Alss+++  +If a wastewater initially containing 300mg/L Cu2+and no dissolved aluminum is dosed with 100mg/L of Al scraps and the above reaction proceeds until equilibrium is reached, what will the final concentrations of Cu2+, Al3+, Cu(s), and Al(s) be?

Answer 1

Final concentrations:

Cu²⁺ = 0

Al³⁺ = 3.13 mmol/L = 84.51 mg/L

Cu = 4.7 mmol/L = 300 mg/L

Al = 0.57 mmol/L = 15.49 mg/L

Step-by-step explanation:

2Al (s) + 3Cu²⁺ (aq) → 2Al³⁺ (aq) + 3Cu (s)

Al: 27 g/mol ∴ 100 mg = 3.7 mmol

Cu: 63.5 g/mol ∴ 300 mg = 4.7 mmol

3 mol Cu²⁺ _______ 2 mol Al

4.7 mmol Cu²⁺ _____ x

x = 3.13 mmol Al

4.7 mmol of Cu²⁺ will be consumed.

3.13 mmol of Al will be consumed.

4.7 mmol of Cu will be produced.

3.13 mmol of Al³⁺ will be produced.

0.57 mmol of Al will remain.