Writing Dissociation Reaction/Equilibrium Expression PLEASE HELP!!

I've attempted this problem multiple times but I keep getting the incorrect answers :( If anyone could help me writing each out and explaining it I would greatly appreciate it!

Write the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (For the dissociation reaction, include states-of-matter under the given conditions in your answer. Use the lowest possible whole number coefficients. Concentration equilibrium expressions take the general form: Kc = [HCl]2 / [H2] . [Cl2].)
(a) HC6H11O2(aq)
dissociation reaction:

equilibrium expression:

(b) Ru(H2O)63+
dissociation reaction:

equilibrium expression:

(c) C3H7NH3+
dissociation reaction:

equilibrium expression:

I'll give you a general approach, and one example. You can do the rest. For an acid-base equilibrium where the equilibrium constant is Ka, the dissociation reaction is always written like this:

Acid (aq) + H2O (l) <---> Conjugate base (aq) + H3O+ (aq)

The expression for Ka is always written this way:

Ka = [Conj base][H3O]+]/[Acid]

Leave out the water. That is the accepted convention for expressing Ka.

So let's do an example for HCl:

HCl (aq) + H2O (l) <---> Cl- (aq) + H3O+ (aq)

Ka = [Cl-][H3O+]/[HCl]

Let's do the third one of yours:

C3H7NH3+ (aq) + H2O (l) <---> C3H7NH2 (aq) + H3O+ {aq)

Ka = [C3H7NH2][H3O+]/[C3H7NH3+]

Your turn. Just a little hint

Conjugate base of HC6H11O2 (aq) is C6H11O2- (aq)
Conjugate base of Ru(H2O)63+ is Ru(H2O)5(OH-)2+