Answer:
5.00 and 2.50 moles of aqueous sodium and sulfide ions are formed.
Explanation:
The dissociation reaction of sodium sulfide is as follows.
[tex]Na_{2}S(aq)\rightarrow 2Na^{+}(aq)+S^{2-}(aq)[/tex]
From the reaction one mole of sodium sulfide produce 2 moles sodium ions.
Let's calculate the moles of [tex]Na^{+}[/tex]ions.
[tex]Moles\,of\,Na^{+}=2.50molNa_{2}S\times \frac{2mol\,Na^{+}}{1mol\,Na_{2}S}=5.00mol\,Na^{+}[/tex]
From the reaction one mole of sodium sulfide produce one mole of sulfide ions.
Let's calculate the moles of [tex]S^{2-}[/tex]ions.
[tex]Moles\,of\,S^{2-}=2.50molNa_{2}S\times \frac{1mol\,S^{2-}}{1mol\,Na_{2}S}=2.50mol\,S^{2-}[/tex]
