95574ul2ze5/20021take/questions/4702807 ssessment is made up of 9 questions that

Question

95574ul2ze5/20021take/questions/4702807 ssessment is made up of 9 questions that may be matching, true/false, numerical entry, or . As this is now the fourth pre-lab assessment you will complete, you are well aware of the ement to pass the pre-lab. You have been given 3 attempts to achieve this. However, I wan rage you to use all 3 attempts even if you pass on the first try. This ensures that you don on the table. There is no extra credit in this course. At the end of the term, your grade will indicate, with minimal rounding. An 89.2% will remain a B: An 89.5% will round up to ies apply so complete all attempts before the due date. imit: There is no time limit for this quiz. If you lose your Internet connection, you may r to the exam where you left off. If you encounter technical difficulties, please contact your Question7 20 pt If you react 3.1g of lead nitrate with 3.7g of potassium iodide, what is the theoretical yield of the lead iodide precipitate? 1 Previous Next

Explanation / Answer

Pb(NO3)2 (aq) + 2KI(aq) --------------> PbI2(s) + 2KNO3(aq)

no of moles of Pb(NO3)2   = W/G.M.Wt

                                           = 3.1/331   = 0.0094 moles

no of moles of KI               = W/G.M.Wt

                                           = 3.7/166   = 0.022 moles

2 moles of KI react with 1 mole of Pb(NO3)2

0.022 moles of KI react with = 1*0.022/2   = 0.011moles of Pb(NO3)2 is required

              Pb(NO3)2 is limiting reactant

1 moles of Pb(NO3)2 react with KI to gives 1 moles of PbI2

0.0094 moles of Pb(NO3)2 react with KI to gives 0.0094 moles of PbI2

mass of PbI2 = no of moles * gram molar mass

                      = 0.0094*461   = 4.34g of PbI2

Theoritical yield of PbI2 = 4.34g

Related Questions

Navigate

• Browse (All)
• Browse 9
• Subjects

---

---

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