Optional extra credit: You are working in an industrial lab and are charged with

Question

Optional extra credit: You are working in an industrial lab and are charged with waste reclamation. The waste that you are working with contains silver chloride and lead (II) Chloride. Both compounds are insoluble in cold water; however, lead (II) chloride is soluble in hot water. Your lab is equipped with beakers, stirring rods, filter paper, a suction filtration apparatus, deionized water, and an oven. Write the chemical formulas for silver chloride and lead (II) chloride. Propose a procedure for separating these two components. After completing the separation, you find out that the percent recovered for one of the components is low. Based on your purposed procedure, briefly explain which component has the low percent recovery and why.

Explanation / Answer

a.
silver chloride : AgCl
lead(II)Chloride = PbCl2

b.
Procedure:
Put some waste in beaker and add deionized water to it.
Start heating the beaker with continuous stirring.
Once it is heated, filter the solution.
PbCl2 will past through and AgCl can be obtained as residue.

c.
PbCl2 might be low because whole of PbCl2 might not have dissolved in water due to less temperature.

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

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