Answer to Question #290771 in General Chemistry for pay

Question

What is the maximum possible mass of lead(II) iodide precipitate that could form from the reaction of 250 mL of 0.25 M lead(II) nitrate mixed with 300 mL of 0.35 M potassium iodide?

Answer 1

Pb(NO3)2 + 2 KI = PbI2 + 2KNO3

Molecular mass of PbI2 = 461

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

300ml of 0.35M KI = 300×0.35=105 m.moles

So 105 × 10^-3 moles of KI will produce 1/2 ×105 m.moles of PbI2

= (105/2000×461)

= 24.2g of PbI2