How many grams of PbBr2 will precipitate when excess CuBr2 solution is added to 68.0 mL of 0.682 M Pb(NO3)2 solution?

CuBr₂(aq) + Pb(NO₃)₂(aq) → PbBr₂(aq) + Cu(NO₃)₂(aq)
1 mole of Pb(NO₃)₂ gives 1 mole of PbBr₂ precipitate.

No. of moles of Pb(NO₃)₂ used = 0.682 × (68.0/1000) = 0.0464 mol
No. of moles of PbBr₂ precipitate = 0.0464 mol

Molar mass of PbBr₂ precipitate = 207.2 + 79.9×2 = 367 g/mol
Mass of PbBr₂ precipitate = 0.0464 × 367 = 17.0 g

367 g/mole X 0.682 mole/L X 0.068 L = 17.02 g