Question

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

Answer 1

Answer:

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

b. $Pb^{2+}(aq)+2I^-(aq)\rightarrow PbI_2(s)$

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.