Question

If a solution containing 30.61 g of lead(II) nitrate is allowed to react completely with a solution containing 5.102 g of sodium sulfide, how many grams of solid precipitate will be formed? mass of solid precipitate: How many grams of the reactant in excess will remain after the reaction? mass of excess reactant: Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (O) for the number of moles. Pb2+ = mol NO3 = Na+ =

Answer 1

mole Imole fear 30.61g Imole semiting Pb(NO3)2 lagg & Nagar Pbs o + 2 Natag + 2 Maze 15.1029 molar mass= 331.2g/mol 70'0452g/mol moles - gram = 30.619 5:1024 Molarmass 331.2g|mos - 78.0450g/mol = 0.0924 mole1 = 00654 mole excess reactant I mole of Mars after reaction produce - Imole pbs 0065 4 " = 0 654 mole pbs mass of Pbs = mole x molar mass af pbs = 0654 mx 239.39/mol = .00 15.64 g) Pbs . I mole Nads react with = Imole Pb(NO3)2 .:.0654 " " " = .0654 1 " Excess mole of PMO) 0924 mole- 0654 mole excess - 0.027 mole Pb(NO₃)2 mass of Pb(NO3)2 = excess mole x molar mass -0.027 mole x 331.2g|mol mass of exces (PHNO3) = 0.9424 8 Pb (NO3)₂ mole of Pb = excess mole of Ph MO3)2 = 0.27 mole, p624 " " NO3 = 2X exsmole of Pb(NO3) = 10924 mole x2 11 Nat = 2x mole of Nass - 1848 mole Noi - 2x 0.065 -0.1300 mole Not - O mole