Question

# Sr(NO3)2(aq)+ 2KlO3(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 KlO3(aq)...

Sr(NO3)2(aq)+ 2KlO3(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 KlO3(aq) (limiting)?

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

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

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

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