How many grams of PbBr2 will precipitate when excess CrBr; solution is added to 60.0 mL of 0.551 M Pb (NO3) 2 solution?

3Pb (NO3) 2 (aq) + 2CrBr3 (aq) - >3PbBrz (s) + 2Cr (NO3) (aq)

5.48 g

Explanation:

M: 367.01

3Pb (NO3) 2 (aq) + 2CrBr3 (aq) → 3PbBr2 (s) + 2Cr (NO3) 3 (aq)

1. Moles of Pb (NO3) 2

n = 30.0 mL * (0.551 mmol/1 mL) = 16.53 mmol Pb (NO3) 2

2. Moles of PbBr2

n = 16.53 mmol Pb (NO3) 2 * (3 mmol PbBr/3 mol Pb (NO3) 2)

= 16.53 mmol PbBr2

(3) Mass of PbBr2 formed

m = 16.53 mmol * (367.01 mg/1 mol) = 6067 mg

(4) Mass of PbBr2 in solution

PbBr2 is slightly soluble, so a significant amount will remain in solution.

Its solubility is 973 mg/100 mL.

Mass of dissolved PbBr2 = 60.0 mL * 973 mg/100 mL = 584 mg

(5) Mass of precipitate

Mass of precipitate = mass of PbBr2 formed - mass of PbBr2 in solution

m = 6067 mg - 584 mg = 5480 mg = 5.48 g

The mass of precipitate is 5.48 g.