[JonathanEyoon]

Huh!? 


0.0634 M?

[AWK]

right side would be Na + S?

wrong!

Na2S dissociates forming cations and anions.
Of course, even that problem can be solved correctly in numbers

[azmanam]

Na2S dissociates forming cations and anions.

True, good point.  Can't forget about the charges,  my bad.

0.0634 M?

Not sure what that is supposed to represent.  show us some work?

If it was a guess at the final answer, then you're off by a factor of two.  Remember, moles of sodium ion in 40.0 mL of a 0.474 M solution is for sodium ion, not sodium sulfide.

[ARGOS++]

Dear JonathanEyoon, Dear Reader;

(JonathanEyoon: Sorry that you ran already the second time ‘against’ a wall!)

If you feel lost, even with such simple Problems, start with the “Stoichiometry Recipe”:
A.)    Reaction:               Na₂S (s)     --->        Na⁺ (aq)    +        S^2- (aq)
B.)    Balancing:           1 Na₂S (s)     --->    2  Na⁺ (aq)    +      1 S^2- (aq)
C₁.)   MW’s:                   78.1 g/m                   27.0 g/m             32.1 g/m   
For this problem even the line: C₁.) is not really required. 

So let’s do the dilution first:
You should remember that 1.0 molar means 1.0 moles per 1.0 Liter Solution, and you have to take only 0.040 Liter, so:    0.474 moles L^-1 * 0.040 L = 0.01896 moles Na₂S.
But this amount is now put into a total volume of 0.300 L final solution, so that it results in a concentration of:  0.01896 moles / 0.300 L   =   0.0632 molar Na₂S.

Finally from line B.) and/or from the Statement in your Reply #13 you know that you get 2  Na⁺ (aq) for each Na₂S, so you should get:  2  *  0.0632 molar  =  0.1264 molar Na⁺.

For the whole “Stoichiometry Recipe” (incl. Excesses) and future explanations you may visit again:   "Stoichiometry Problem”.

I hope it may be of help to you all.

Good Luck!
                    ARGOS⁺⁺