Question

Write a balanced molecular and net ionic equation for the following reactions

a. insoluble aluminum hydroxide reacts with hydrobromic acid
b. Lead (I) nitrate reacts with lithium iodide to produce a precipitate and a soluble compound.

Answer 1

a.
`Al(OH)_3(s)+3H^\rightarrow Al^(3+)+3H_2O(l)\\`

b.
`Pb^(2+)(aq)+2I^-(aq)\rightarrow PbI_2(s)`

Step-by-step explanation:

Hello,

a. In this case, the overall reaction is:

`Al(OH)_3(s)+3HBr(aq)\rightarrow AlBr_3(aq)+3H_2O`

Nevertheless, the ionic version is:

`Al(OH)_3(s)+3H^++3Br^-(aq)\rightarrow Al^(3+)+3Br^-(aq)+3H_2O(l)\\`

Since the base is insoluble, thereby, the balanced net ionic equation turns out:

`Al(OH)_3(s)+3H^\rightarrow Al^(3+)+3H_2O(l)\\`

Since bromide ions become spectator ions.

b) In this case, the overall reaction is:

`Pb(NO_3)_2(aq)+2LiI(aq)\rightarrow PbI_2(s)+2LiNO_3(aq)`

Nevertheless, the ionic version is:

`Pb^(2+)(aq)+2(NO_3)^-(aq)+2Li^+(aq)+2I^-(aq)\rightarrow PbI_2(s)+2Li^+(aq)+2(NO_3)^-(aq)`

Since lead (II) iodide is insoluble whereas lithium nitrate does not, thereby, the net ionic equation turns out:

`Pb^(2+)(aq)+2I^-(aq)\rightarrow PbI_2(s)`

Since lithium and nitrate ions become spectator ions.

Regards.