Question

How many grams of PbBr2 will precipitate when excess AlBr3 solution is added to 45.0 mL of 0.745 M Pb(NO3)2 solution?

3Pb(NO3)2(aq) + 2AlBr3(aq) 3PbBr2(s) + 2Al(NO3)3(aq)

______g

Answer 1

Ans:

Amount of Pb(NO3)2 in given solution = Molecular weight of Pb(NO3)2 × 45 ML ×0.745 Molar ÷ 1000

molecular weight of Pb(NO3)2 = 331.208

Hence amount of Pb(NO3)2 in gm = 331.208 × 45 × 0.745 ÷ 1000

Hence amount of Pb(NO3)2 = 11.103 gm

Content of Pb in Pb(NO3)2 is calculated as

molar mass of Pb(NO3)2 = molar mass of Pb

331.208. = 207.977

11.103 gm of Pb(NO3)2 = ? gm of Pb

Therefore Pb Content = 11.103 × 207.977 ÷ 331.208

Pb in solution = 6.972 gm

Now content of Pb(Br)2 in solution is calculated as

Molar mass of Pb. = Molar mass of PbBr2

207.977 = 367.008

6.972 gm of Pb =. ? gm of PbBr2

Now content of PbBr2 gm in Solution = 6.972 × 367.008 ÷ 207.977

= 12.303gm

Hence 12.303 gm PbBr2 will precipitate out from above solution