Question

Sr(NO₃)_2(aq)+ 2KlO_3(aq)--------->Sr(IO₃)_2(S)+2K⁺_(aq)+2NO₃^-_(aq)

a) How many moles of Sr(IO₃)_2(aq) will be formes from 5mL of 0.10M KlO_3(aq) (limiting)?

b)Write the net ionic equation associated with this reaction.

c)Write the Ksp expression associated with the precipitate in this reaction.

Answer 1

a) First we need to calculate mol of KIO3

mol KIO3= 0.1M x 5x10^-3L= 5x10^-4 mol

If we lok at the balanced equation we can see that 2 mol of KIO3 produces 1 mol Sr(IO3)2, so, 5x10^-4 mol of KIO3 produce 5x10^-4/2 mol of Sr(IO3)2= 2.5x10^-4 mol.

b) To write the net ionic equation we have to dissociate all compount into it`s ions and then eliminate the spectator ions. Spectator ions are those who are at both sides of the equation. You need to know that the solid compounds can`t be dissociated. Let´s do it with our equation:

Sr+2 + 2NO3- + 2K+ + 2IO3- -------> Sr(IO3)2(s) + 2K+ + 2NO3-

In this case NO3- and K+ are the spectator ions, they are at both sides. So, we just need to eliminate them to obtain the net ionic equation:

Sr+2 + 2IO3- -------> Sr(IO3)2(s)

c) Ksp= [Sr+2][IO3-]²