How many grams of PbI₂ will precipitate from the reaction of 14.0 mL of 0.190 M Pb(NO₃)₂ with excess KI solution? Assume that all of the PbI₂ is insoluble.

Question

Quizplus · Accepted Answer

First, calculate the moles of Pb(NO3)2 using its concentration and volume: moles = 0.190 M * 0.014 L = 0.00266 moles. Since the reaction is 1:1 for Pb(NO3)2 to PbI2, the moles of PbI2 will also be 0.00266. The molar mass of PbI2 is approximately 461 g/mol, so the mass of PbI2 precipitated is 0.00266 moles * 461 g/mol = 1.23 g.

How Many Grams of PbI2 Will Precipitate from the Reaction

How Many Grams of PbI₂ Will Precipitate from the Reaction